Chapter 11 Three Dimensional Geometry (Class 12 - Maths NCERT Concept Notes)

Direction Cosines and Direction Ratios of a Line

In 3-dimensional geometry, the orientation of a line is defined by the angles it makes with the three mutually perpendicular coordinate axes. To understand this, we must first distinguish between a Position Vector and a Free Vector.

Position Vector vs. Free Vector

A Free Vector, such as $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$, is characterized only by its magnitude and direction; its location in space is not fixed. However, for mathematical convenience, we can always represent this free vector as a Position Vector $\vec{OP}$ by shifting it such that its tail is at the origin $O(0,0,0)$ and its head is at the point $P(a, b, c)$.

Because the direction of the line segment $OP$ is identical to the original free vector, we can study the direction of any line in space by analyzing a parallel line passing through the origin.

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Direction Cosines (DCs) of a Line

Consider a line $AB$ located anywhere in 3D space. To define its direction analytically, we draw a line $QP$ parallel to $AB$ that passes through the origin $O$. This directed line $OP$ is used as the reference to define the direction of $AB$.

1. Direction Angles

The ray $OP$ makes specific angles with the positive directions of the $X, Y,$ and $Z$ axes. These angles are denoted by the Greek letters $\alpha$, $\beta$, and $\gamma$.

• $\alpha$: Angle with the positive $X$-axis ($OX$).

• $\beta$: Angle with the positive $Y$-axis ($OY$).

• $\gamma$: Angle with the positive $Z$-axis ($OZ$).

These three angles are collectively known as the Direction Angles of the line $AB$.

2. Defining Direction Cosines

The cosine values of these direction angles are called the Direction Cosines of the line. They are typically represented by the letters $l$, $m$, and $n$.

The set of direction cosines is written as $\langle l, m, n \rangle$.

The Distinction: Line vs. Directed Ray

A crucial distinction in 3D geometry is that while a vector or a ray has a single unique direction, a line extends infinitely in two opposite directions. Consequently, every line in space is associated with two sets of direction cosines.

Case 1: The Ray $OP$ (Direction $A \to B$)

If the line is directed from $A$ towards $B$, it makes angles $\alpha, \beta, \gamma$ with the axes. The direction cosines are:

Case 2: The Ray $OQ$ (Direction $B \to A$)

If we reverse the direction, the line now makes angles with the negative directions of the axes. Geometrically, these new angles are the supplements of the original angles:

• Angle with $X$-axis = $(\pi - \alpha)$

• Angle with $Y$-axis = $(\pi - \beta)$

• Angle with $Z$-axis = $(\pi - \gamma)$

Derivation for Reversed DCs:

Using the trigonometric identity $\cos(\pi - \theta) = -\cos \theta$, we calculate the new direction cosines:

Conclusion

Therefore, the direction cosines of a line are not unique in sign. They are given as:

Remember: if the question asks for the DCs of a directed line segment (vector), provide one set. If it asks for the DCs of a line, both $\langle l, m, n \rangle$ and $\langle -l, -m, -n \rangle$ are technically correct, though usually the positive set is used for further calculations.

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Direction Cosines of the Coordinate Axes

The coordinate axes—namely the X-axis, Y-axis, and Z-axis—are the most fundamental lines in 3D geometry. Since they are mutually perpendicular, we can easily determine their direction cosines by analyzing the angles they make with each other.

1. Direction Cosines of the X-axis

To find the direction cosines of the X-axis, we look at the angles ($\alpha, \beta, \gamma$) it makes with the three coordinate axes:

• It makes an angle of $0^\circ$ with the positive X-axis ($\alpha = 0^\circ$).

• It is perpendicular to the Y-axis, so it makes an angle of $90^\circ$ with it ($\beta = 90^\circ$).

• It is perpendicular to the Z-axis, so it makes an angle of $90^\circ$ with it ($\gamma = 90^\circ$).

Calculation:

Thus, the direction cosines of the X-axis are $\langle 1, 0, 0 \rangle$.

2. Direction Cosines of the Y-axis

Similarly, for the Y-axis, the angles are $\alpha = 90^\circ$, $\beta = 0^\circ$, and $\gamma = 90^\circ$:

Thus, the direction cosines of the Y-axis are $\langle 0, 1, 0 \rangle$.

3. Direction Cosines of the Z-axis

For the Z-axis, the angles are $\alpha = 90^\circ$, $\beta = 90^\circ$, and $\gamma = 0^\circ$:

Thus, the direction cosines of the Z-axis are $\langle 0, 0, 1 \rangle$.

Summary Table of Axis Direction Cosines

Coordinate Axis	$l$ ($\cos \alpha$)	$m$ ($\cos \beta$)	$n$ ($\cos \gamma$)
X-axis	$1$	$0$	$0$
Y-axis	$0$	$1$	$0$
Z-axis	$0$	$0$	$1$

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Relation Between Direction Cosines

The direction cosines of a line are not independent of each other. There exists a fundamental quadratic relationship between $l, m,$ and $n$ that stems from the 3D distance formula.

Theorem

If $l, m, n$ are the direction cosines of a line, then the sum of their squares is always equal to unity (1).

Derivation

Given: Let $L$ be a line with direction cosines $\langle l, m, n \rangle$. Through the origin $O$, draw a line $OP$ parallel to $L$. Let the coordinates of point $P$ be $(x, y, z)$ and the distance $OP$ be $r$.

Construction: From point $P$, draw perpendiculars $PA, PB, PC$ to the $X, Y,$ and $Z$ axes respectively to complete a rectangular box. Thus, $OA = x$, $OB = y$, and $OC = z$.

Proof:

Using the distance formula in 3D space, the distance $r$ from origin to $P(x, y, z)$ is given by:

In $\Delta OAP$, the angle $\angle OAP = 90^\circ$ (since $PA \perp OX$) and $\angle AOP = \alpha$ (direction angle). Therefore:

Since $l = \cos \alpha$, we have:

Similarly, by considering triangles $\Delta OBP$ and $\Delta OCP$, we get:

Now, substituting the values of $x, y,$ and $z$ from equations (ii) and (iii) into equation (i):

Dividing both sides by $r^2$ (since $r \neq 0$):

Coordinates in terms of DCs

The coordinates of any point $P$ at a distance $r$ from the origin on a line with direction cosines $l, m, n$ are given by:

Answer:

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Direction Ratios of a Line

While Direction Cosines (DCs) provide a unique way to describe the orientation of a line (subject to sign), they often involve square roots and fractions. To simplify calculations, we use Direction Ratios (DRs), which are any three numbers proportional to the actual direction cosines of the line.

If $\langle l, m, n \rangle$ are the direction cosines of a line $L$ and $\lambda$ is any non-zero real number, then the set of numbers $\langle \lambda l, \lambda m, \lambda n \rangle$ are called the Direction Ratios or Direction Numbers of the line $L$.

Thus, if $\langle a, b, c \rangle$ are the direction ratios, then:

Since $\lambda$ can be any non-zero real number, a single line can have infinitely many sets of direction ratios.

To Find Actual Direction Cosines from Direction Ratios

Given the direction ratios $\langle a, b, c \rangle$ of a line, we can derive the unique direction cosines $\langle l, m, n \rangle$ using the fundamental relation $l^2 + m^2 + n^2 = 1$.

Derivation:

From the definition of direction ratios, we have the proportionality:

This implies:

We know that for any set of direction cosines:

Substituting the values from equation (i):

Substituting this value of $k$ back into equation (i), we obtain the actual direction cosines:

Key Comparison: DCs vs. DRs

Feature	Direction Cosines (DCs)	Direction Ratios (DRs)
Uniqueness	Unique for a directed line (fixed values).	Infinite sets possible for a single line.
Relationship	Sum of squares is always 1 ($l^2+m^2+n^2=1$).	Sum of squares is not usually 1 ($a^2+b^2+c^2 \neq 1$).
Notation	Typically $\langle l, m, n \rangle$.	Typically $\langle a, b, c \rangle$.

Fundamental Remarks

1. Any set of three numbers proportional to $l, m, n$ can serve as direction ratios. For example, if DCs are $\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle$, then $\langle 1, 1, 1 \rangle$ or $\langle 2, 2, 2 \rangle$ are valid direction ratios.

2. If any of the direction ratios are zero, the corresponding direction cosine is also zero. For instance, if $a = 0$, then $l = 0$.

3. Direction ratios are also known as Direction Numbers.

Answer:

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Direction Cosines and Ratios of a Line Joining Two Points

In three-dimensional space, exactly one line passes through two distinct given points. We can determine the orientation of this line by calculating its direction cosines and direction ratios based on the coordinates of these two points.

Geometric Derivation

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two points in space. Let $l, m, n$ be the direction cosines of the line $PQ$, making angles $\alpha, \beta, \gamma$ with the positive $X, Y$ and $Z$ axes respectively.

1. Setup and Construction

From points $P$ and $Q$, draw perpendiculars $PM$ and $QN$ to the $XY$-plane. From $P$, draw a perpendicular $PL$ to the line segment $QN$.

In the resulting configuration, the line segment $PL$ is parallel to the $XY$-plane, and the triangle $PQL$ is a right-angled triangle, with the right angle at $L$.

2. Deriving the Component along Z-axis

The length of the segment $LQ$ is the difference between the $Z$-coordinates of $Q$ and $P$.

In the right-angled triangle $\Delta PQL$, the angle $\angle PQL$ is equal to the direction angle $\gamma$ (the angle the line makes with the $Z$-axis).

3. Generalizing for X and Y axes

By drawing similar perpendiculars to the $YZ$ and $ZX$ planes, we can derive the other two direction cosines:

Summary of Formulas

Direction Cosines (DCs)

The direction cosines of the line joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are:

Where $PQ$ is the distance between the two points:

Direction Ratios (DRs)

Since direction ratios are any three numbers proportional to direction cosines, we can multiply the DCs by the constant $PQ$. This gives us the Direction Ratios of the line segment $PQ$:

Geometric Significance and Projections

To understand why these formulas work, we can visualize the projections of the line segment $PQ$ onto the coordinate axes.

1. Projection on X-axis: The length of the shadow of $PQ$ on the X-axis is $|x_2 - x_1|$. Since $\cos \alpha = \frac{\text{Projection}}{\text{Actual Length}}$, it naturally follows that $l = \frac{x_2 - x_1}{PQ}$.

2. The Relation $l^2 + m^2 + n^2 = 1$: If you square and add the three direction cosines derived above:

This confirms that the formulas derived from two points are mathematically consistent with the properties of direction cosines.

Effect of Direction

The order of points matters when determining the sign of DRs and DCs.

• For a line directed from $P$ to $Q$, the DRs are $\langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$.

• For a line directed from $Q$ to $P$, the DRs are $\langle x_1 - x_2, y_1 - y_2, z_1 - z_2 \rangle$.

This results in the two possible sets of DCs: $\pm \langle l, m, n \rangle$. For a line (which has no fixed direction), either set is acceptable. For a vector $\vec{PQ}$, only the first set is correct.

Answer:

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Angle Between Two Lines

In two-dimensional geometry, two lines are either intersecting or parallel. However, in three-dimensional space, we encounter a third possibility: Skew Lines. These are lines that are neither parallel nor intersecting and do not lie in the same plane (non-coplanar).

To define and calculate the angle between any two lines in space (whether they are coplanar or skew), we use the concept of parallel translation.

Defining the Angle between Non-Coplanar (Skew) Lines

Let $L_1$ and $L_2$ be two skew lines in space. Since they do not intersect, we cannot measure the angle between them directly. The angle is defined as follows:

1. Take any arbitrary point in space, say $A$. In practice, the Origin ($O$) is usually chosen for simplicity.

2. Through point $A$, draw two lines $L'_1$ and $L'_2$ such that $L'_1 \parallel L_1$ and $L'_2 \parallel L_2$.

3. The angle between the intersecting lines $L'_1$ and $L'_2$ is defined as the angle between the original lines $L_1$ and $L_2$.

Geometric Observations

1. Existence of Two Angles

When two lines intersect (like $L'_1$ and $L'_2$), they actually form two pairs of vertically opposite angles. This results in two possible angles between the lines:

• An angle $\theta$ (usually the acute angle).

• Its supplementary angle $\pi - \theta$ (usually the obtuse angle).

Generally, when we refer to "the angle" between two lines, we refer to the acute angle ($\theta$), unless specified otherwise.

2. Case of Parallel Lines

If the two given lines $L_1$ and $L_2$ are parallel, they have the same orientation in space. When translated to a common point $A$, the lines $L'_1$ and $L'_2$ will coincide.

3. Use of the Origin (Remark)

The most convenient way to find these angles is to draw the parallel lines through the Origin ($O$). By doing so, the angles that the original line makes with the coordinate axes are identical to the angles made by the parallel line passing through the origin. This allows us to use Direction Cosines and Direction Ratios directly to find the angle.

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Angle Between Two Lines whose Direction Cosines are Given

When the orientation of two lines $L_1$ and $L_2$ is defined by their direction cosines, we can use trigonometric relationships to find the angle $\theta$ between them. This analytical approach relies on the properties of vectors and the distance formula in three-dimensional space.

Analytical Derivation

Setup: Let $L_1$ and $L_2$ be two lines with direction cosines $\langle l_1, m_1, n_1 \rangle$ and $\langle l_2, m_2, n_2 \rangle$ respectively. Through the origin $O$, draw two lines $OP$ and $OQ$ parallel to $L_1$ and $L_2$.

By property of parallel lines, the angle $\theta$ between $L_1$ and $L_2$ is the same as the angle between $OP$ and $OQ$.

Let us take points $P$ and $Q$ on these lines such that $OP = 1$ and $OQ = 1$. The coordinates of $P$ and $Q$ will be:

• $P(l_1, m_1, n_1)$

• $Q(l_2, m_2, n_2)$

Step 1: Using the Distance Formula

In $\Delta OPQ$, the square of the distance $PQ$ is given by:

Expanding the squares:

$PQ^2 = (l_2^2 + l_1^2 - 2l_1l_2) + (m_2^2 + m_1^2 - 2m_1m_2) + (n_2^2 + n_1^2 - 2n_1n_2)$

Rearranging the terms:

$PQ^2 = (l_1^2 + m_1^2 + n_1^2) + (l_2^2 + m_2^2 + n_2^2) - 2(l_1l_2 + m_1m_2 + n_1n_2)$

Since the sum of squares of direction cosines is 1 ($l^2 + m^2 + n^2 = 1$):

Step 2: Using the Cosine Rule

To determine the relationship between the distance $PQ$ and the angle $\theta$, we apply the Law of Cosines (also known as the Cosine Rule) to the triangle formed in three-dimensional space.

In any triangle, the Law of Cosines relates the lengths of the sides to the cosine of one of its angles. It acts as an extension of the Pythagorean Theorem for non-right-angled triangles. For a triangle with sides $a, b, c$ and an angle $C$ opposite to side $c$, the formula is expressed as:

Application to Triangle $OPQ$

In our derivation, we consider the triangle $\Delta OPQ$ where:

• Side $OP$ is the distance from the origin to point $P$.

• Side $OQ$ is the distance from the origin to point $Q$.

• Side $PQ$ is the distance between the two points on the respective lines.

• $\theta$ is the angle subtended at the origin $O$ between the vectors $\vec{OP}$ and $\vec{OQ}$.

According to the standard setup for direction cosines, we specifically choose points $P$ and $Q$ such that they are at a unit distance from the origin. This simplification is mathematically sound because direction cosines are independent of the magnitude of the vectors.

Substituting these values into the Cosine Rule for $\Delta OPQ$:

Final Simplification of the Expression

By replacing the lengths with $1$, the equation reduces to a very simple form involving only the cosine of the angle:

$PQ^2 = (1)^2 + (1)^2 - 2(1)(1) \cos \theta$

$PQ^2 = 1 + 1 - 2 \cos \theta$

Step 3: Equating the Results

Comparing the value of $PQ^2$ obtained in step 1 and 2, we get:

$2 - 2 \cos \theta = 2 - 2(l_1l_2 + m_1m_2 + n_1n_2)$

Cancelling 2 from both sides and dividing by $-2$:

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Derivation of the Sine Formula for Angle Between Two Lines

To find $\sin \theta$ when we already know $\cos \theta$, we utilize the fundamental Pythagorean identity in trigonometry:

Rearranging this to solve for $\sin^2 \theta$:

Utilizing Direction Cosine Properties

We know that for any line, the sum of the squares of its direction cosines is always unity. Thus, for two lines $L_1$ and $L_2$:

• $(l_1^2 + m_1^2 + n_1^2) = 1$

• $(l_2^2 + m_2^2 + n_2^2) = 1$

Multiplying these two identities together, we still get $1$. We can therefore rewrite the $1$ in the above equation as the product of these two sums:

Application of Lagrange's Identity

To simplify the expression above, we use Lagrange's Identity. In algebra, this identity states that for any two sets of real numbers $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$:

$(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) - (a_1b_1 + a_2b_2 + a_3b_3)^2$

$= (a_1b_2 - a_2b_1)^2 + (a_2b_3 - a_3b_2)^2 + (a_3b_1 - a_1b_3)^2$

By substituting $l, m, n$ for $a$ and $b$, we apply this identity to our equation for $\sin^2 \theta$:

Final Expression for $\sin \theta$

Taking the square root on both sides of equation (ii), we arrive at the final result:

Geometric Significance

This expression is equivalent to the magnitude of the cross product of two unit vectors $\hat{u_1}$ and $\hat{u_2}$ representing the directions of the two lines.

If the lines are parallel, $\theta = 0$, which means $\sin \theta = 0$. For the sum of squares to be zero, each individual term must be zero:

Hence, for parallel lines, the direction cosines are proportional:

Condition of Perpendicularity

If the two lines are perpendicular to each other, the angle $\theta = 90^\circ$.

Substituting this in the formula for cosine derived above, we get:

Condition of Parallelism

When two lines $L_1$ and $L_2$ are parallel in three-dimensional space, they maintain the same orientation relative to the coordinate axes. This geometric relationship has specific implications for their direction cosines (DCs).

Geometric Interpretation

If two lines are parallel, the angle $\theta$ between them is either $0^\circ$ (if they are in the same direction) or $180^\circ$ (if they are in opposite directions). In both scenarios, the lines are considered parallel because they share the same line of action when translated to the origin.

For these angles, the trigonometric values are:

• If $\theta = 0^\circ$, then $\cos \theta = 1$.

• If $\theta = 180^\circ$, then $\cos \theta = -1$.

Therefore, for parallel lines, we must have $|\cos \theta| = 1$.

Mathematical Derivation via Sine Formula

A more robust way to prove the condition of parallelism is using the Sine Formula derived in the previous section. For parallel lines:

Substituting this into the sine expression:

For the square root of a sum of squares to be zero, each individual term must be zero simultaneously:

From equations (i), (ii), and (iii), we conclude that the direction cosines are proportional:

However, we know that for direction cosines, $l_1^2 + m_1^2 + n_1^2 = 1$ and $l_2^2 + m_2^2 + n_2^2 = 1$. Substituting $l_1 = kl_2$, $m_1 = km_2$, and $n_1 = kn_2$:

$k^2(l_2^2 + m_2^2 + n_2^2) = 1$

$k^2(1) = 1 \implies k = \pm 1$

Final Conclusion

Two lines are parallel if and only if their direction cosines are either identical or exactly opposite in sign:

OR

Summary of Relationships

Relationship	Mathematical Condition
General Angle ($\theta$)	$\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$
Perpendicular ($\theta = 90^\circ$)	$l_1l_2 + m_1m_2 + n_1n_2 = 0$
Parallel ($\theta = 0^\circ$)	$l_1 = l_2, m_1 = m_2, n_1 = n_2$

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Angle Between Two Lines whose Direction Ratios are Given

While direction cosines provide a normalized description of a line's orientation, Direction Ratios (DRs) are far more common in practical coordinate geometry and vector algebra. Direction ratios are any three numbers $a, b, c$ that are proportional to the direction cosines $l, m, n$. They often involve integers or simpler fractions, making manual calculations more efficient.

Analytical Derivation of the Cosine Formula

If $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ are the direction ratios of two lines $L_1$ and $L_2$ respectively, they relate to the direction cosines as follows:

To find the angle $\theta$ between these lines, we leverage the existing formulas for direction cosines by first performing a conversion.

Step 1: Conversion of DRs to DCs

Given: Direction ratios of $L_1$ as $(a_1, b_1, c_1)$ and $L_2$ as $(a_2, b_2, c_2)$.

The direction cosines for the first line are given by:

Similarly, for the second line $L_2$:

Step 2: Substitution into the Fundamental Formula

From our previous derivation, we know that the angle $\theta$ in terms of direction cosines is:

Substituting the converted values of $l, m, n$ into the above equation:

$\cos \theta = \left(\frac{a_1}{\sqrt{\sum a_1^2}}\right)\left(\frac{a_2}{\sqrt{\sum a_2^2}}\right) + \left(\frac{b_1}{\sqrt{\sum a_1^2}}\right)\left(\frac{b_2}{\sqrt{\sum a_2^2}}\right) + $$ \left(\frac{c_1}{\sqrt{\sum a_1^2}}\right)\left(\frac{c_2}{\sqrt{\sum a_2^2}}\right)$

Step 3: The General Formula for $\cos \theta$

On simplifying the common denominator, we obtain the standard expression used in 3D geometry:

Derivation of the Sine Formula using DRs

Similar to the cosine formula, we can express $\sin \theta$ in terms of direction ratios. This is particularly useful when calculating the vector product or the area of a parallelogram defined by two lines.

Application of Lagrange's Identity with DRs

By substituting $l = \frac{a}{\sqrt{\sum a^2}}$ into the sine formula $\sin \theta = \sqrt{\sum (l_1m_2 - l_2m_1)^2}$, the denominator becomes the product of the magnitudes of the two direction ratio sets.

The General Formula for $\sin \theta$ is expressed as:

Condition of Perpendicularity for Two Lines

In three-dimensional geometry, two lines are said to be perpendicular (or orthogonal) if the angle between them is exactly $90^\circ$. When working with Direction Ratios (DRs), this geometric relationship simplifies into a powerful algebraic identity that is widely used in vector calculus and 3D coordinate geometry.

Recall the general formula for the cosine of the angle $\theta$ between two lines $L_1$ and $L_2$ having direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ respectively:

For perpendicular lines, the angle $\theta = 90^\circ$. We know from trigonometric properties that:

Substituting this value into the general formula, we get:

Final Condition of Orthogonality

Since the denominator in the above expression represents the product of the magnitudes of the two direction sets, it cannot be zero for any properly defined line. Therefore, for the entire fraction to equal zero, the numerator must be zero.

Condition of Parallelism for Two Lines

When two lines $L_1$ and $L_2$ are parallel in three-dimensional space, they point in the same (or exactly opposite) direction. This implies that their Direction Ratios (DRs) must be proportional to one another, meaning one set of DRs is simply a scalar multiple of the other.

To derive this condition rigorously, we use the sine formula for the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:

For parallel lines, the angle $\theta$ is either $0^\circ$ or $180^\circ$. In both cases:

Substituting this into the formula, the entire numerator must equal zero:

Step-by-Step Algebraic Simplification

For the square root of a sum of squares to be zero, every individual squared term must be zero simultaneously because squares of real numbers can never be negative.

Final Proportionality Condition

Combining the results from equations (i), (ii), and (iii), we arrive at the necessary and sufficient condition for two lines to be parallel:

Summary Table: DR Relationships

Relationship	Mathematical Condition
Acute Angle ($\theta$)	$\cos \theta = \left| \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{\sum a_1^2} \cdot \sqrt{\sum a_2^2}} \right|$
Perpendicular ($\perp$)	$a_1a_2 + b_1b_2 + c_1c_2 = 0$
Parallel ($\parallel$)	$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Answer:

Straight Line (in Space)

In three-dimensional geometry, a straight line is a fundamental geometric object. Its position and direction in space can be uniquely determined if we know either:

1. A point on the line and its direction (parallel to a given vector).
2. Two distinct points through which the line passes.

We can represent the equation of a straight line in space in two primary forms: the Vector Form and the Cartesian Form.

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Equation of a Straight Line Passing Through a Given Point and Parallel to a Given Vector

In three-dimensional geometry, a line is uniquely determined if it passes through a specific fixed point and follows a particular direction. This direction is typically represented by a vector parallel to the line.

Vector Equation of the Line

Setup: Let a line $L$ pass through a given point $A$ with position vector $\vec{a}$ and let it be parallel to a given non-zero vector $\vec{b}$. Let $P$ be any arbitrary point on the line with position vector $\vec{r}$.

Since the point $P$ lies on the line passing through $A$, the vector $\vec{AP}$ must be parallel to the vector $\vec{b}$. According to the property of collinear/parallel vectors, there exists a real number $\lambda$ (called a parameter) such that:

We can express the displacement vector $\vec{AP}$ in terms of position vectors using the triangle law of addition as $\vec{OP} - \vec{OA}$. Substituting this in the above equation:

$\vec{r} - \vec{a} = \lambda \vec{b}$

On rearranging the terms, we arrive at the standard vector form:

Cartesian (Symmetrical) Form of the Line

To express the line in the Cartesian coordinate system, we resolve the vectors into their rectangular components.

Analytical Derivation

Let the coordinates of the fixed point $A$ be $(x_1, y_1, z_1)$ and the direction ratios (direction numbers) of the line be $\langle a, b, c \rangle$. For any arbitrary point $P(x, y, z)$ on the line, the vectors are defined as:

• $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

• $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$

• $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$

Substituting these component values into the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$:

$x\hat{i} + y\hat{j} + z\hat{k} = (x_1 + a\lambda)\hat{i} + (y_1 + b\lambda)\hat{j} + (z_1 + c\lambda)\hat{k}$

By equating the coefficients of the base vectors $\hat{i}, \hat{j}, \text{ and } \hat{k}$ from both sides, we obtain the parametric form:

From equations (i), (ii), and (iii), we solve for $\lambda$:

$\lambda = \frac{x - x_1}{a}, \quad \lambda = \frac{y - y_1}{b}, \quad \lambda = \frac{z - z_1}{c}$

Equating these values, we get the Symmetrical Form (or Cartesian Form) of the equation of the line:

Answer:

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Equation of a Straight Line Passing Through Two Given Points

In three-dimensional space, a line is uniquely determined if it passes through two distinct points. This orientation is defined by the displacement vector between these two points, which serves as the direction of the line.

Vector Equation of the Line

Setup: Let a line pass through two given points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$ respectively. Let $P$ be any arbitrary point on the line with position vector $\vec{r}$.

If point $P$ lies on the line passing through $A$ and $B$, then the vector $\vec{AP}$ must be collinear with the vector $\vec{AB}$. By the definition of collinearity, there exists a scalar parameter $\lambda$ such that:

Using the triangle law of vector addition, we can write:

• $\vec{AP} = \vec{OP} - \vec{OA} = \vec{r} - \vec{a}$

• $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$

Substituting these into the equation provided above:

$\vec{r} - \vec{a} = \lambda(\vec{b} - \vec{a})$

By rearranging the terms, we obtain the Vector Equation of the Line:

Cartesian (Symmetrical) Form of the Line

To convert the vector equation into coordinates, we define the components of the given points.

Analytical Derivation

Let the coordinates of point $A$ be $(x_1, y_1, z_1)$ and point $B$ be $(x_2, y_2, z_2)$. Let the arbitrary point $P$ be $(x, y, z)$. The vectors are represented as:

• $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

• $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$

• $\vec{b} - \vec{a} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

Substituting these values into $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$:

$x\hat{i} + y\hat{j} + z\hat{k} = [x_1 + \lambda(x_2 - x_1)]\hat{i} + [y_1 + \lambda(y_2 - y_1)]\hat{j} + $$ [z_1 + \lambda(z_2 - z_1)]\hat{k}$

On equating the coefficients of $\hat{i}, \hat{j}, \text{ and } \hat{k}$, we get the parametric form of the line:

From equations (i), (ii), and (iii), we solve for $\lambda$:

$\lambda = \frac{x - x_1}{x_2 - x_1}, \quad \lambda = \frac{y - y_1}{y_2 - y_1}, \quad \lambda = \frac{z - z_1}{z_2 - z_1}$

Equating these gives the Symmetrical Form (Cartesian Form):

Answer:

Angle Between Two Lines

The angle between two lines is fundamentally defined as the angle between the vectors that represent their directions. Since lines in three-dimensional space can be translated without changing their orientation, the angle $\theta$ depends solely on their direction vectors and is independent of their positions (base points) in space.

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Angle between Two Lines in Vector Form

In three-dimensional space, the orientation of a line is entirely determined by its direction vector. When we represent lines in vector form, we focus on the vectors that provide the "heading" of the lines, rather than their starting positions.

Consider two lines $L_1$ and $L_2$ defined by the equations:

In these representations, the vectors $\vec{a}_1$ and $\vec{a}_2$ are position vectors of fixed points on the lines. However, to calculate the angle $\theta$ between the lines, these position vectors are irrelevant. We focus exclusively on the vectors $\vec{b}_1$ and $\vec{b}_2$, which are parallel to the lines $L_1$ and $L_2$ respectively.

Application of the Scalar (Dot) Product

From the fundamentals of Vector Algebra, we know that the dot product of any two vectors $\vec{A}$ and $\vec{B}$ is related to the cosine of the angle between them by the formula $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$. Applying this to our direction vectors $\vec{b}_1$ and $\vec{b}_2$:

To find the angle, we rearrange the equation above to solve for $\cos \theta$:

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Angle Between Two Lines in Cartesian Form

In 3D coordinate geometry, lines are most frequently expressed in their symmetrical (Cartesian) form. Calculating the angle between lines in this format involves extracting their Direction Ratios (DRs) and applying the principles of vector dot products to these components.

Consider two lines $L_1$ and $L_2$ passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ respectively. Their equations are given by:

Here, the denominators $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ represent the Direction Ratios of the two lines. These ratios are essential as they are directly proportional to the components of the vectors parallel to the lines.

To find the angle $\theta$, we treat the direction ratios as components of two direction vectors $\vec{b}_1$ and $\vec{b}_2$:

• $\vec{b}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$

• $\vec{b}_2 = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$

The cosine of the angle between these vectors is defined by the Scalar Product formula:

Expanding the Numerator (Dot Product)

The dot product of the two direction vectors is the sum of the products of their corresponding direction ratios:

Expanding the Denominator (Magnitudes)

The magnitude of each direction vector is the square root of the sum of the squares of its direction ratios:

• $|\vec{b}_1| = \sqrt{a_1^2 + b_1^2 + c_1^2}$

• $|\vec{b}_2| = \sqrt{a_2^2 + b_2^2 + c_2^2}$

The General Cartesian Formula

By substituting these expanded terms back into the cosine relationship, we obtain the standard formula used for solving coordinate geometry problems:

Note: The modulus sign in the numerator $|...|$ is used to ensure that the value of $\cos \theta$ is positive, thereby resulting in an acute angle $\theta$.

Practical Application and Convention

When using this formula, it is a common convention in mathematics to provide the angle $\theta$ as an inverse cosine function if the value is not a standard trigonometric ratio:

Using Direction Cosines

If instead of direction ratios, we are given the Direction Cosines (DCs) $\langle l_1, m_1, n_1 \rangle$ and $\langle l_2, m_2, n_2 \rangle$ for the two lines, the formula simplifies significantly.

Since the sum of squares of direction cosines is always unity ($l^2 + m^2 + n^2 = 1$), the denominator in the general formula becomes $1 \cdot 1$.

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Conditions for Particular Orientations

In three-dimensional space, the relative orientation of two lines is determined by the relationship between their Direction Ratios (DRs). The two extreme cases of these orientations are when the lines are either Perpendicular (orthogonal) or Parallel (collinear direction).

1. Condition of Perpendicularity ($\theta = 90^\circ$)

Two lines $L_1$ and $L_2$ with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if the angle between them is $90^\circ$. This orientation is critical in problems involving the Shortest Distance or Foot of the Perpendicular.

Analytical Derivation

From the general cosine formula, we know that the angle $\theta$ depends on the dot product of the direction vectors. For perpendicular lines:

Substituting this into the Cartesian cosine formula:

Since the denominator represents lengths (magnitudes) and cannot be zero, the numerator must vanish. This leads to the Orthogonality Condition:

2. Condition of Parallelism ($\theta = 0^\circ$ or $180^\circ$)

Two lines are parallel if their direction vectors point in the same or exactly opposite directions. In such cases, the direction ratios of one line are a scalar multiple of the other.

Analytical Derivation via Sine Formula

When lines are parallel, the angle $\theta = 0^\circ$, and therefore $\sin 0^\circ = 0$. Using the sine formula derived from Lagrange's Identity:

For the numerator to be zero, each squared term must be zero individually:

• $a_1b_2 - a_2b_1 = 0 \implies \frac{a_1}{a_2} = \frac{b_1}{b_2}$

• $b_1c_2 - b_2c_1 = 0 \implies \frac{b_1}{b_2} = \frac{c_1}{c_2}$

• $c_1a_2 - c_2a_1 = 0 \implies \frac{c_1}{c_2} = \frac{a_1}{a_2}$

Combining these, we obtain the Condition of Proportionality (or Collinearity Condition):

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Equation of a Line Passing Through a Given Point and Perpendicular to Two Given Lines

Determining the equation of a line that is simultaneously perpendicular to two other lines is a common problem in 3D geometry. This relies on the property of the vector cross product, which allows us to find a direction that is orthogonal to two known directions.

Vector Equation of the Required Line

Given:

1. A fixed point through which the line passes, with position vector $\vec{a}$.

2. Two existing lines in space:

Analytical Derivation

Since the required line is perpendicular to both $L_1$ and $L_2$, its direction must be perpendicular to both direction vectors $\vec{b}_1$ and $\vec{b}_2$. From vector algebra, we know that the cross product of two vectors is perpendicular to both of them.

Therefore, the direction vector $\vec{m}$ of the required line is parallel to:

Substituting this direction into the standard line equation $\vec{r} = \vec{a} + t\vec{m}$, we get the Vector Equation:

Cartesian Form Representation

To find the equation in Cartesian form, we need to calculate the Direction Ratios (DRs) of the required line by expanding the cross product $\vec{b}_1 \times \vec{b}_2$.

Determination of Direction Ratios

Let $\vec{b}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{b}_2 = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$. The cross product is given by the determinant:

Expanding the determinant, the direction ratios $\langle a, b, c \rangle$ of the required line are:

• $a = (b_1c_2 - b_2c_1)$

• $b = (c_1a_2 - c_2a_1)$

• $c = (a_1b_2 - a_2b_1)$

If the given point is $(x_1, y_1, z_1)$, the Cartesian Equation is:

Answer:

Shortest Distance Between Two Lines

In three-dimensional geometry, the relative position of two lines can be categorized into three types: intersecting, parallel, or skew lines. Calculating the shortest distance between them is a fundamental concept used to determine if lines meet or how far apart they remain in space.

Understanding Skew Lines

In a two-dimensional plane, any two lines are either parallel or they must intersect at some point. However, in three-dimensional space, a third possibility exists. Two straight lines which are neither parallel nor intersecting are called skew lines.

Skew lines do not lie in the same plane (they are non-coplanar). Because they are not parallel, they have different directions, and because they do not intersect, they never meet regardless of how far they are extended.

Line of Shortest Distance

For any two skew lines $L_1$ and $L_2$, there exists one and only one line that is perpendicular to both $L_1$ and $L_2$. This unique line is known as the line of shortest distance.

If we identify a point $M$ on line $L_1$ and a point $N$ on line $L_2$ such that the segment $MN$ is perpendicular to both lines, then the length of this segment $MN$ represents the minimum possible distance between any point on $L_1$ and any point on $L_2$.

Definition of Shortest Distance (S.D.)

Let $L_1$ and $L_2$ be two lines in space. Let $MN$ be a line segment which intersects $L_1$ at point $M$ and $L_2$ at point $N$ such that $MN$ is at right angles to both $L_1$ and $L_2$. The length of this segment $MN$ is called the Shortest Distance (abbreviated as S.D.) between the two lines.

Mathematical Significance of S.D.

The shortest distance serves as a test for the intersection of lines:

Remark: Two non-parallel lines in 3D space intersect if and only if the shortest distance between them is zero. If the S.D. is non-zero and the lines are not parallel, they are confirmed to be skew lines.

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The Shortest Distance Between Two Non-Parallel Lines

For two non-parallel and non-intersecting lines (skew lines) in three-dimensional space, the shortest distance is the length of a segment that is perpendicular to both lines simultaneously. This distance is calculated by finding the projection of a vector joining any two points on the lines onto the direction of their common perpendicular.

Vector Derivation of Shortest Distance

The calculation of the shortest distance between two skew lines (lines that are neither parallel nor intersecting) is a centerpiece of three-dimensional vector geometry. The derivation relies on the geometric concept of projections and the properties of the vector cross product.

Geometric Logic of the Derivation

Consider two skew lines $L_1$ and $L_2$. Any segment joining a point on $L_1$ to a point on $L_2$ will have a specific length. However, the shortest of all such segments is the one that is perpendicular to both lines. Let this segment be $MN$.

To find the length of $MN$, we pick any two arbitrary points $A_1$ (on $L_1$) and $A_2$ (on $L_2$). The segment $MN$ is essentially the projection of the vector $\vec{A_1A_2}$ onto the line that is perpendicular to both $L_1$ and $L_2$.

Establishing the Direction of Shortest Distance

The lines are given by the vector equations:

The directions of these lines are defined by vectors $\vec{b}_1$ and $\vec{b}_2$. A vector that is simultaneously perpendicular to both directions is obtained using the Cross Product:

To use this for projection, we convert it into a unit vector $\hat{n}$ by dividing it by its magnitude:

Applying the Projection Formula

The vector joining the two points $A_1(\vec{a}_1)$ and $A_2(\vec{a}_2)$ is:

The shortest distance $d$ is the magnitude of the scalar projection of $\vec{A_1A_2}$ along the unit vector $\hat{n}$:

Substituting the values of $\vec{A_1A_2}$ and $\hat{n}$:

Final Vector Result

By rearranging the scalar multiplication, we get the standard formula for the Shortest Distance (S.D.):

Crucial Points for Interpretation

1. Modulus Sign: The modulus $|\dots|$ is essential because distance is always a non-negative scalar quantity, regardless of the orientation of the cross product.

2. Dot Product Significance: The numerator $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$ is actually the Scalar Triple Product, which geometrically represents the volume of a parallelepiped formed by these three vectors. The distance $d$ is therefore the altitude of this parallelepiped.

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Cartesian Form of Shortest Distance

While the vector form of the shortest distance is theoretically elegant, the Cartesian Form is the most practical for manual calculations during examinations. This form translates vector operations—specifically the Scalar Triple Product and the Vector Cross Product—into coordinate-based determinants and algebraic expressions.

Understanding the Numerator: The Determinant

In the vector formula, the numerator is the absolute value of the scalar triple product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$. In Cartesian coordinates, where point $A_1 = (x_1, y_1, z_1)$ and point $A_2 = (x_2, y_2, z_2)$, the vector $(\vec{a}_2 - \vec{a}_1)$ is given by:

$(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

The direction ratios of the two lines are $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$. The entire scalar triple product can be computed directly using a 3x3 Determinant:

Understanding the Denominator: Magnitude of Cross Product

The denominator of the shortest distance formula is the magnitude of the cross product of the two direction vectors, $|\vec{b}_1 \times \vec{b}_2|$. To find this, we first find the cross product vector using a determinant:

Expansion of this determinant gives:

$(b_1c_2 - b_2c_1)\hat{i} - (a_1c_2 - a_2c_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$

The magnitude (length) of this resulting vector is the square root of the sum of the squares of its components:

The Comprehensive Symmetrical Formula

By combining the absolute value of the determinant and the magnitude from the above equations, we arrive at the final Cartesian Formula for Shortest Distance ($d$):

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Condition for the Intersection of Two Lines

In three-dimensional geometry, two non-parallel lines are said to intersect if they lie in the same plane (are coplanar) and the shortest distance between them is zero. If the shortest distance is non-zero, the lines are categorized as skew lines.

Intersection in Vector Form

Given: Two non-parallel lines $L_1$ and $L_2$ represented by:

Condition of Intersection

The given lines $L_1$ and $L_2$ will intersect if and only if the Shortest Distance (S.D.) between them is exactly zero. Since the formula for S.D. is $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$, for $d = 0$, the numerator must be zero.

Geometric Interpretation

Above equation implies that the vectors $(\vec{a}_2 - \vec{a}_1)$, $\vec{b}_1$, and $\vec{b}_2$ are coplanar. If this condition is satisfied and the vectors $\vec{b}_1$ and $\vec{b}_2$ are not parallel, the lines must meet at a single point.

Intersection in Cartesian Form

Given: Two non-parallel lines $L_1$ and $L_2$ in symmetrical form:

Necessary and Sufficient Condition

The lines will intersect if the determinant formed by the difference of coordinates and the direction ratios vanishes:

This condition ensures that the point-to-point vector and the two direction vectors lie in the same 3D plane.

Skew Lines

Skew lines are lines that are non-parallel and non-intersecting. This relationship is defined by the negation of the intersection conditions.

Vector Condition for Skew Lines

Two lines are skew if their scalar triple product is non-zero:

Cartesian Condition for Skew Lines

Two lines are skew if the intersection determinant is non-zero:

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Distance Between Two Parallel Lines

In three-dimensional space, two lines are parallel if they have the same direction (or proportional direction ratios) and do not intersect. Unlike skew lines, parallel lines are always coplanar, meaning a single plane can be found that contains both lines. The distance between them is constant at every point.

Vector Equation and Setup

Given: Let the two parallel lines $L_1$ and $L_2$ be defined by the following vector equations:

Notice that both lines share the same direction vector $\vec{b}$. Let $A_1$ be a point on line $L_1$ with position vector $\vec{a}_1$, and $A_2$ be a point on line $L_2$ with position vector $\vec{a}_2$.

Geometric Construction

Let $N$ be the foot of the perpendicular drawn from point $A_2$ to the line $L_1$. The distance $d$ between the lines is the length of the segment $A_2N$.

Analytical Derivation

Let $\theta$ be the angle between the vector $\vec{A_1A_2}$ and the direction vector $\vec{b}$. In the right-angled triangle $\Delta A_1NA_2$, we have:

We know that $\vec{A_1A_2} = \vec{a}_2 - \vec{a}_1$. Now, consider the cross product of vector $\vec{b}$ and vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = |\vec{b}| |\vec{a}_2 - \vec{a}_1| \sin \theta \, \hat{n}$

Where $\hat{n}$ is a unit vector perpendicular to the plane containing the lines. Taking the magnitude on both sides:

$|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = |\vec{b}| |\vec{a}_2 - \vec{a}_1| \sin \theta$

Final Substitution

Substituting the value from equation (iii), where $|\vec{a}_2 - \vec{a}_1| \sin \theta = d$:

By rearranging the above equation, we obtain the formula for the Distance between Parallel Lines:

Cartesian Implementation

If the equations are given in Cartesian form, the lines will have the same denominators $(a, b, c)$ but different points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. To calculate the distance:

1. Identify vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

2. Find vector $\vec{a}_2 - \vec{a}_1 = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.

3. Compute the cross product and divide by the magnitude of $\vec{b}$.

Answer:

Plane

In the study of Three Dimensional Geometry, a plane is defined as a surface such that if any two distinct points are selected on it, the entire straight line passing through these points resides completely within that surface. This implies that every point constituting the line is a part of the plane.

A fundamental characteristic of a plane is its relationship with perpendicular lines: if a specific line is perpendicular to a plane, then it is necessarily perpendicular to every single line contained within that plane.

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Ways to Determine a Unique Plane

A particular plane can be uniquely identified or "specified" using several geometric criteria. While there are numerous methods, the following six are primary:

(i) Via Three Non-Collinear Points: A unique plane exists that passes through any three points, provided they do not lie on the same straight line.

(ii) Via Two Concurrent Lines: One and only one plane can be constructed that contains two lines intersecting at a single common point.

(iii) Via Two Parallel Lines: A single unique plane can be drawn to encompass two lines that are parallel to one another.

(iv) Via Normal and Distance from Origin: A plane is uniquely determined if we know its perpendicular distance from the origin and the direction of the normal (a vector perpendicular to the plane).

(v) Via a Point and a Normal Direction: A unique plane is specified by a single point through which it passes and a fixed direction to which it is perpendicular.

(vi) Via a Point and Two Parallel Lines: One and only one plane can be established that passes through a specific point and remains parallel to two given lines.

Summary Table of Specifications

Specification Type	Conditions for Uniqueness
Point-based	Three points must be non-collinear.
Line-based (Concurrent)	Two lines must intersect at one point.
Line-based (Parallel)	Two lines must be parallel to each other.
Normal-based	Requires distance from origin or a fixed point.
Analytical Preference	Methods (iv) and (v) are most useful for vector equations.

It is important to note that among the methods listed above, those involving the normal vector and distance (iv and v) are highly significant because they lead to the most simplified and standard versions of the vector equation of a plane.

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Plane Perpendicular to a Given Direction and at a Given Distance from the Origin

A plane can be uniquely defined if its orientation (perpendicular direction) and its specific distance from the origin are known. This is often referred to as the Normal Form of the plane equation.

Vector Equation of the Plane

Consider a plane $P_1$ located at a perpendicular distance $p$ from the origin $O$. Let $\hat{n}$ represent a unit vector that is perpendicular to the plane, directed away from the origin. Let $N$ be the foot of the perpendicular drawn from the origin to the plane, such that the vector $\vec{ON} = p\hat{n}$.

Let $P$ be any arbitrary point in the plane with a position vector $\vec{r}$. The point $P$ lies on the plane if and only if the vector $\vec{NP}$ is perpendicular to the unit normal vector $\hat{n}$. Since $\hat{n}$ is perpendicular to the entire plane, it must be perpendicular to every line segment lying within that plane.

Derivation of the Vector Form

Based on the condition of perpendicularity, the dot product of $\vec{NP}$ and $\hat{n}$ must be zero:

We can express $\vec{NP}$ as the difference between the position vectors of $P$ and $N$:

$\vec{NP} = \vec{r} - p\hat{n}$

Substituting this into the perpendicularity condition:

$(\vec{r} - p\hat{n}) \cdot \hat{n} = 0$

$\vec{r} \cdot \hat{n} - p(\hat{n} \cdot \hat{n}) = 0$

Above equation represents the vector equation of the plane in normal form, where $p$ is the length of the perpendicular from the origin.

Cartesian Equation of the Plane

To convert the vector equation into Cartesian coordinates, let the direction cosines of the normal vector $\vec{ON}$ be $l, m,$ and $n$. Thus, the unit normal vector is:

$\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$

Let the coordinates of any point $P$ on the plane be $(x, y, z)$, so its position vector is:

$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

Substituting these into the vector equation $\vec{r} \cdot \hat{n} = p$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (l\hat{i} + m\hat{j} + n\hat{k}) = p$

This linear equation is the Cartesian equation of the plane in Normal Form.

Standard and General Forms

While the Normal Form is mathematically elegant, it is often more practical to represent a plane using general vectors or linear equations. This leads us to the Standard Vector Form and the General Cartesian Form.

Standard Vector Form

In the normal form, we specifically use a unit vector $\hat{n}$. However, in most geometric problems, we are given a general normal vector $\vec{n}$ which may not have a magnitude of $1$.

Let $\vec{n}$ be any vector perpendicular to the plane, and let $p$ be the perpendicular distance of the plane from the origin. We know from the Normal Form that:

Since $\hat{n}$ is a unit vector in the direction of $\vec{n}$, we can define it as:

Substituting this value of $\hat{n}$ into equation provided above, we get:

$\vec{r} \cdot \left( \frac{\vec{n}}{|\vec{n}|} \right) = p$

$\vec{r} \cdot \vec{n} = p |\vec{n}|$

Since both $p$ (distance) and $|\vec{n}|$ (magnitude) are constants, their product is also a constant. Let this constant be $d$.

Note: In this form, $d$ represents $p|\vec{n}|$. Therefore, $d$ is not the distance from the origin unless the magnitude of $\vec{n}$ is unity. To find the actual distance $p$ from the origin, we use:

$p = \frac{|d|}{|\vec{n}|}$

General Cartesian Form

Every first-degree equation in three variables $x, y,$ and $z$ represents a plane. This is known as the General Form of a plane's equation.

Let the direction ratios (direction numbers) of the normal vector $\vec{n}$ be $A, B,$ and $C$. Thus:

$\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$

Let $\vec{r}$ be the position vector of any point $P(x, y, z)$ on the plane:

$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

Substituting these into the standard vector equation $\vec{r} \cdot \vec{n} = d$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (A\hat{i} + B\hat{j} + C\hat{k}) = d$

$Ax + By + Cz = d$

To write it in the most general form, we move all terms to one side:

$Ax + By + Cz - d = 0$

Let $D = -d$. The equation becomes:

This is the General Cartesian Equation of a plane. In this equation, the coefficients $A, B,$ and $C$ are the Direction Ratios of the normal to the plane. In Indian mathematical terminology, these are frequently referred to as the Attitude Numbers of the plane.

Conversion: General Form to Normal Form

To convert a general equation $Ax + By + Cz + D = 0$ into the normal form $lx + my + nz = p$, we follow these steps:

1. Shift the constant term to the right side: $Ax + By + Cz = -D$.

2. Ensure the right-hand side is positive (since distance $p$ cannot be negative). If $-D$ is negative, multiply the entire equation by $-1$.

3. Divide the entire equation by the magnitude of the normal vector, which is $\sqrt{A^2 + B^2 + C^2}$.

Comparison of Plane Equations

Form Type	Equation	Key Variables
Vector (Normal)	$\vec{r} \cdot \hat{n} = p$	$\hat{n}$ is unit vector; $p$ is distance.
Cartesian (Normal)	$lx + my + nz = p$	$l, m, n$ are direction cosines.
Vector (Standard)	$\vec{r} \cdot \vec{n} = d$	$\vec{n}$ is any normal vector.
Cartesian (General)	$Ax + By + Cz + D = 0$	$A, B, C$ are direction numbers.

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Plane Passing Through a Given Point and Perpendicular to a Given Direction

A plane can be uniquely determined if we know a specific point through which it passes and the direction that is perpendicular to its surface. This is a very common way to represent a plane in Three-Dimensional Geometry, often referred to as the One-Point Form.

Vector Form

Let a plane $P_1$ pass through a fixed point $A$ with position vector $\vec{a}$. Let $\vec{n}$ be a given vector that is perpendicular (normal) to the plane.

Let $P$ be any arbitrary point on the plane with position vector $\vec{r}$. The vector $\vec{AP}$ lies entirely within the plane. Since the vector $\vec{n}$ is perpendicular to the plane, it must be perpendicular to every vector in that plane, including $\vec{AP}$.

Derivation of Vector Equation

The condition for point $P$ to lie in the plane is:

Mathematically, if two vectors are perpendicular, their dot product is zero:

We can express $\vec{AP}$ in terms of position vectors as $\vec{AP} = \vec{r} - \vec{a}$. Substituting this into the above equation, we get:

Cartesian Form

To find the Cartesian equivalent, we represent the points and the normal vector in terms of their coordinates and direction numbers.

Given:

Fixed point $A = (x_1, y_1, z_1)$

Arbitrary point $P = (x, y, z)$

Direction numbers (Attitude numbers) of $\vec{n} = \langle A, B, C \rangle$

Derivation of Cartesian Equation

The vectors can be written as:

$\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$

$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

$\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$

Therefore, the vector $\vec{r} - \vec{a}$ is:

$\vec{r} - \vec{a} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$

Applying the dot product $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$:

$[(x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}] \cdot [A\hat{i} + B\hat{j} + C\hat{k}] = 0$

By solving the dot product, we obtain the One-Point Form of the plane:

Answer:

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Position of a Line with Respect to a Plane

In three-dimensional space, a line can occupy three distinct positions relative to a given plane: it can lie entirely within the plane, it can be parallel to the plane (never intersecting), or it can intersect the plane at exactly one point.

Vector Approach

To understand how a line and a plane interact in three-dimensional space, we analyze their vector equations. The intersection of a line and a plane depends entirely on the relationship between the direction of the line and the normal of the plane.

The Mathematical Framework

Let the straight line be represented by its vector equation:

Here, $\vec{a}$ is the position vector of a fixed point on the line, and $\vec{b}$ is the direction vector parallel to the line.

Let the plane be represented by its vector equation:

Here, $\vec{n}$ is the normal vector (perpendicular to the surface) and $d$ is a scalar constant.

Any point on the line has the general position vector $\vec{r} = \vec{a} + \lambda\vec{b}$. If this line interacts with the plane, there must be a value of $\lambda$ such that this vector satisfies the plane's equation. By substituting equation (i) into equation (ii), we obtain:

$(\vec{a} + \lambda\vec{b}) \cdot \vec{n} = d$

$\vec{a} \cdot \vec{n} + (\lambda\vec{b}) \cdot \vec{n} = d$

Rearranging this to solve for $\lambda$:

The behavior of the line relative to the plane is determined by the nature of the solution for $\lambda$ in this linear equation.

Analysis of Geometric Cases

Case (i): The Line Lies Completely in the Plane

If the line resides entirely within the plane, every point on the line must satisfy the plane's equation. This means equation (iii) must be true for all real values of $\lambda$. This is only possible if the equation becomes an identity ($0 = 0$).

Conditions:

1. $\vec{b} \cdot \vec{n} = 0$: The line's direction is perpendicular to the plane's normal (making it parallel to the surface).

2. $\vec{a} \cdot \vec{n} = d$: The fixed point $\vec{a}$ of the line actually lies on the plane.

Case (ii): The Line is Parallel to the Plane

A line is parallel to a plane if it never intersects it. In this case, equation (iii) must have no solution for $\lambda$. This occurs when the coefficient of $\lambda$ is zero, but the constant term on the right is non-zero.

Conditions:

1. $\vec{b} \cdot \vec{n} = 0$: The line is parallel to the surface of the plane.

2. $\vec{a} \cdot \vec{n} \neq d$: The fixed point $\vec{a}$ of the line does not lie on the plane.

Case (iii): The Line Meets the Plane at a Unique Point

If the line is not parallel to the plane, it must pierce the plane at exactly one point. This happens when equation (iii) yields a unique value for $\lambda$.

Condition:

1. $\vec{b} \cdot \vec{n} \neq 0$: The line's direction is not perpendicular to the normal vector. This means the line is "inclined" relative to the plane.

The unique value of $\lambda$ is calculated as:

$\lambda = \frac{d - \vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}}$

Summary of Vector Interactions

Geometric Relation	Mathematical Condition	Interpretation
Coincident (Line in Plane)	$\vec{b} \cdot \vec{n} = 0$ AND $\vec{a} \cdot \vec{n} = d$	Direction is parallel and point is on plane.
Parallel (No Intersection)	$\vec{b} \cdot \vec{n} = 0$ AND $\vec{a} \cdot \vec{n} \neq d$	Direction is parallel but point is off plane.
Intersecting (One Point)	$\vec{b} \cdot \vec{n} \neq 0$	Line is inclined towards the plane.

Cartesian Approach

In Three-Dimensional Geometry, the relationship between a line and a plane can be analyzed using their Cartesian equations. By substituting the parametric coordinates of a general point on the line into the equation of the plane, we can determine the nature of their intersection.

Mathematical Framework

Consider a straight line passing through the point $(x_1, y_1, z_1)$ with direction ratios $a, b,$ and $c$. Its Symmetrical Form is given by:

Consider the General Equation of a plane with direction numbers (attitude numbers) $A, B,$ and $C$:

From equation (i), any arbitrary point $P$ on the line can be expressed in terms of the parameter $\lambda$ as follows:

If this point $P$ lies on the plane, its coordinates must satisfy the plane equation (ii). Substituting these values:

$A(x_1 + a\lambda) + B(y_1 + b\lambda) + C(z_1 + c\lambda) + D = 0$

Expanding and regrouping the terms to separate the parameter $\lambda$:

$(Aa\lambda + Bb\lambda + Cc\lambda) + (Ax_1 + By_1 + Cz_1 + D) = 0$

Detailed Analysis of Geometric Conditions

The geometric relationship between the line and the plane depends on the number of solutions for $\lambda$ in equation (iii).

1. Condition for a Line to Lie Entirely in the Plane

If every single point of the line resides within the plane, the line is said to be coincident with the plane. This implies that equation (iii) must be an identity, meaning it is true for every real value of $\lambda$. This occurs if:

AND

2. Condition for a Line to be Parallel to the Plane

A line is parallel to a plane if they never intersect, regardless of how far they are extended. Mathematically, this means equation (iii) has no solution for $\lambda$. This occurs when the coefficient of $\lambda$ is zero, but the constant part is non-zero:

$\mathbf{Aa + Bb + Cc = 0}$ (The line is perpendicular to the plane's normal)

AND

$\mathbf{Ax_1 + By_1 + Cz_1 + D \neq 0}$ (The point on the line is not on the plane)

3. Condition for a Line to Intersect at a Unique Point

The line will pierce the plane at exactly one unique point if equation (iii) yields a single, specific value for $\lambda$. This is possible only when the coefficient of $\lambda$ is non-zero:

In this case, the intersection point can be found by solving for $\lambda$:

$\lambda = -\frac{Ax_1 + By_1 + Cz_1 + D}{Aa + Bb + Cc}$

Summary of Cartesian Conditions

Relationship	Constraint on Direction ($a, b, c$)	Constraint on Point ($x_1, y_1, z_1$)
Line lies in the Plane	$Aa + Bb + Cc = 0$	$Ax_1 + By_1 + Cz_1 + D = 0$
Line parallel to the Plane	$Aa + Bb + Cc = 0$	$Ax_1 + By_1 + Cz_1 + D \neq 0$
Line intersects the Plane	$Aa + Bb + Cc \neq 0$	No specific constraint

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Summary Table of Conditions

Relationship	Vector Condition	Cartesian Condition
Line lies in Plane	$\vec{b} \cdot \vec{n} = 0$ and $\vec{a} \cdot \vec{n} = d$	$Aa+Bb+Cc=0$ and $Ax_1+By_1+Cz_1+D=0$
Line parallel to Plane	$\vec{b} \cdot \vec{n} = 0$ and $\vec{a} \cdot \vec{n} \neq d$	$Aa+Bb+Cc=0$ and $Ax_1+By_1+Cz_1+D \neq 0$
Unique Intersection	$\vec{b} \cdot \vec{n} \neq 0$	$Aa+Bb+Cc \neq 0$

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Plane Passing Through Two Given Points and Parallel to a Given Line

In three-dimensional space, a unique plane can be determined if it passes through two distinct points and maintains a parallel orientation to a specified line. This configuration effectively provides three constraints: two points on the plane and one direction vector that is parallel to the surface of the plane.

Vector Form

Given:

1. Two points $A$ and $B$ on the plane with position vectors $\vec{a}_1$ and $\vec{a}_2$ respectively.

2. A line with equation $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{b}$ is the direction vector of the line.

Derivation

Let $P$ be any arbitrary point on the plane with position vector $\vec{r}$. We consider three vectors related to the plane:

1. Vector $\vec{AP} = \vec{r} - \vec{a}_1$ (lies in the plane).

2. Vector $\vec{AB} = \vec{a}_2 - \vec{a}_1$ (lies in the plane).

3. Vector $\vec{b}$ (parallel to the plane).

For the point $P$ to lie in the plane, the vectors $\vec{AP}$, $\vec{AB}$, and $\vec{b}$ must be coplanar. The mathematical condition for three vectors to be coplanar is that their scalar triple product must be equal to zero.

Substituting the position vectors:

Cartesian Form

To express the equation in Cartesian coordinates, we translate the vectors into their component forms.

Coordinates:

Point $A = (x_1, y_1, z_1)$

Point $B = (x_2, y_2, z_2)$

Direction ratios of the parallel line $\vec{b} = \langle a, b, c \rangle$

General point $P = (x, y, z)$

Derivation

The components of the vectors are:

$\vec{r} - \vec{a}_1 = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

$\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$

The scalar triple product from the vector form equation can be represented as a determinant. Thus, the Cartesian equation of the plane is:

Answer:

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Plane Passing Through a Given Point and Parallel to Two Given Lines

A specific plane is uniquely identified if it passes through a known fixed point and is oriented such that it remains parallel to two non-parallel lines. In this configuration, the direction of the plane is governed by the direction vectors of the two given lines.

Vector Form

Given:

1. A fixed point $A$ on the plane with position vector $\vec{a}$.

2. Two non-parallel lines represented by $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{d}$, where $\vec{b}$ and $\vec{d}$ are the direction vectors of these lines.

Derivation of the Equation

Let $P$ be any arbitrary point on the required plane with position vector $\vec{r}$. We can construct the vector $\vec{AP}$ as follows:

$\vec{AP} = \vec{r} - \vec{a}$

Since the plane is parallel to the vectors $\vec{b}$ and $\vec{d}$, and the vector $\vec{AP}$ lies entirely within the plane, these three vectors ($\vec{AP}$, $\vec{b}$, and $\vec{d}$) must be coplanar. For three vectors to be coplanar, their scalar triple product must vanish.

Substituting the position vector of point $A$ into the equation:

This equation can also be written in the form $\vec{r} \cdot (\vec{b} \times \vec{d}) = \vec{a} \cdot (\vec{b} \times \vec{d})$.

Cartesian Form

To express this in Cartesian coordinates, we define the points and the direction ratios of the parallel lines.

Let:

Fixed point $A = (x_1, y_1, z_1)$

Direction ratios of the first line $\vec{b} = \langle a_1, b_1, c_1 \rangle$

Direction ratios of the second line $\vec{d} = \langle a_2, b_2, c_2 \rangle$

General point $P = (x, y, z)$

Determinant Representation

The vector $\vec{AP}$ is $(x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$. The scalar triple product from equation (i) can be converted into a determinant form:

Answer:

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Condition of Coplanarity of Two Lines

In three-dimensional geometry, two lines are said to be coplanar if they lie in the same plane. For two lines to be coplanar, they must either be parallel or they must intersect. If they are neither parallel nor intersecting, they are termed as skew lines.

Vector Form

Let us consider two lines in their vector representations:

Observation:

1. The first line passes through point $A$ with position vector $\vec{a}$ and is parallel to vector $\vec{b}$.

2. The second line passes through point $C$ with position vector $\vec{c}$ and is parallel to vector $\vec{d}$.

Derivation of the Condition

The vector joining the points $A$ and $C$ is given by:

$\vec{AC} = \text{Position Vector of } C - \text{Position Vector of } A = \vec{c} - \vec{a}$

For the two lines to be coplanar, the vector $\vec{AC}$ must be coplanar with the direction vectors $\vec{b}$ and $\vec{d}$. This means the Scalar Triple Product of these three vectors must be zero.

Cartesian Form

Let the two lines be given in their symmetric Cartesian forms:

From the equations:

Line (i) passes through $A(x_1, y_1, z_1)$ and has direction ratios $\langle a_1, b_1, c_1 \rangle$.

Line (ii) passes through $C(x_2, y_2, z_2)$ and has direction ratios $\langle a_2, b_2, c_2 \rangle$.

Determinant Condition

The condition $(\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d}) = 0$ can be expressed in determinant form using the coordinates and direction ratios:

Answer:

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Plane Containing Two Lines

A single unique plane can contain two straight lines if and only if those lines are coplanar. This means the lines must either intersect at a common point or be parallel to each other. If the lines are skew, no such plane exists.

Vector Form

Consider two coplanar lines represented by the following vector equations:

Analysis of Vectors:

1. The first line passes through point $A$ with position vector $\vec{a}$ and is parallel to vector $\vec{b}$.

2. The second line passes through point $C$ with position vector $\vec{c}$ and is parallel to vector $\vec{d}$.

Derivation

Let $P$ be any arbitrary point on the plane with position vector $\vec{r}$. For point $P$ to reside on the plane that contains both lines, the following three vectors must be coplanar:

i. $\vec{AP} = \vec{r} - \vec{a}$

ii. $\vec{AC} = \vec{c} - \vec{a}$

iii. $\vec{b}$ (The direction vector of the first line)

Since these vectors lie in the same plane, their scalar triple product must be zero:

Substituting the position vectors into the condition:

Cartesian Form

To derive the Cartesian equation, we represent the points and direction ratios in coordinate geometry terms.

Line 1: Passes through $A(x_1, y_1, z_1)$ with direction numbers $\langle a_1, b_1, c_1 \rangle$.

Line 2: Passes through $C(x_2, y_2, z_2)$ with direction numbers $\langle a_2, b_2, c_2 \rangle$.

Determinant Equation

Let $P(x, y, z)$ be a general point on the plane. The vectors $\vec{AP}$, $\vec{AC}$, and $\vec{b}$ are expressed as:

$\vec{AP} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$

$\vec{AC} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

$\vec{b} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$

Applying the condition of coplanarity via a determinant:

Important Note:

The above determinant results in a linear equation in $x, y,$ and $z$, representing the required plane. For this plane to exist, the lines must not be skew lines.

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Plane Containing Two Parallel Lines

When two lines are parallel, they share the same or proportional direction ratios. To define a unique plane containing such lines, we cannot use the direction vectors of both lines (as they are the same). Instead, we utilize a point from each line to construct a second vector lying within the plane.

Vector Form

Consider two parallel lines represented as:

Observations:

1. Line (i) passes through point $A$ with position vector $\vec{a}$.

2. Line (ii) passes through point $C$ with position vector $\vec{c}$.

3. Both lines are parallel to the same direction vector $\vec{b}$.

Derivation

Let $P$ be any general point on the required plane with position vector $\vec{r}$. To find the equation, we identify two vectors that lie in the plane:

i. Vector $\vec{AP} = \vec{r} - \vec{a}$

ii. Vector $\vec{AC} = \vec{c} - \vec{a}$

Since the plane contains both parallel lines, it must also be parallel to the direction vector $\vec{b}$. Thus, the vectors $(\vec{r} - \vec{a})$, $(\vec{c} - \vec{a})$, and $\vec{b}$ are coplanar. The scalar triple product of these vectors must be zero:

In this equation, the cross product $(\vec{c} - \vec{a}) \times \vec{b}$ provides the normal vector to the plane.

Cartesian Form

Let the two parallel lines be represented in Cartesian form as follows:

Parameters:

Point $A = (x_1, y_1, z_1)$ and Point $C = (x_2, y_2, z_2)$

Direction ratios for both lines = $\langle a, b, c \rangle$

Determinant Formulation

For any point $P(x, y, z)$ on the plane, the vectors $\vec{AP}$, $\vec{AC}$, and the direction vector $\vec{b}$ are coplanar. This leads to the following determinant equation:

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Plane Passing Through Three Points

A specific plane can be uniquely determined if it passes through three non-collinear points. If the points were collinear (lying on a single straight line), infinitely many planes could pass through them, rotating around that line like pages of a book.

Vector Form

Let three given points be $A, B,$ and $C$ with position vectors $\vec{a}, \vec{b},$ and $\vec{c}$ respectively. Let $P$ be any arbitrary point on the required plane with the position vector $\vec{r}$.

Derivation

From the given points, we can define three vectors that must lie within the plane:

1. $\vec{AP} = \vec{r} - \vec{a}$

2. $\vec{AB} = \vec{b} - \vec{a}$

3. $\vec{AC} = \vec{c} - \vec{a}$

For the point $P$ to lie in the same plane as $A, B,$ and $C$, the vectors $\vec{AP}, \vec{AB},$ and $\vec{AC}$ must be coplanar. Mathematically, the scalar triple product of these three vectors must be equal to zero.

Substituting the position vectors into the above equation, we obtain the vector equation of the plane:

Cartesian Form

To convert the vector equation into Cartesian coordinates, let the three points be:

$A = (x_1, y_1, z_1)$

$B = (x_2, y_2, z_2)$

$C = (x_3, y_3, z_3)$

And let the general point be $P = (x, y, z)$.

Determinant Representation

The components of the three coplanar vectors are:

$\vec{AP} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$

$\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

$\vec{AC} = (x_3 - x_1)\hat{i} + (y_3 - y_1)\hat{j} + (z_3 - z_1)\hat{k}$

The scalar triple product from the vector equation is most conveniently expressed as a $3 \times 3$ determinant. Thus, the Cartesian equation of the plane is:

Answer:

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Intercept Form

The Intercept Form is a specific way of representing the equation of a plane based on the lengths it cuts off from the coordinate axes. This form is particularly useful when the points where the plane meets the $x, y,$ and $z$ axes are known.

Consider a plane that does not pass through the origin. Let this plane intersect the $x$-axis at $A$, the $y$-axis at $B$, and the $z$-axis at $C$. The distances of these points from the origin are known as the intercepts.

Let:

$OA = a$ (x-intercept)

$OB = b$ (y-intercept)

$OC = c$ (z-intercept)

Thus, the coordinates of the points are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.

Derivation of the Intercept Form

Let the general equation of the required plane be:

Since the plane passes through the point $A(a, 0, 0)$, this point must satisfy the above equation the above (i):

This gives us: $Aa = -D$ or $A = -\frac{D}{a}$.

Similarly, since the plane passes through $B(0, b, 0)$ and $C(0, 0, c)$:

Now, substituting the values of $A, B,$ and $C$ back into equation (i):

$\left(-\frac{D}{a}\right)x + \left(-\frac{D}{b}\right)y + \left(-\frac{D}{c}\right)z + D = 0$

Dividing the entire equation by $-D$ (where $D \neq 0$ because the plane does not pass through the origin):

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0$

Rearranging the terms, we get the standard Intercept Form:

Answer:

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Intersection of Two Planes

When two non-parallel planes meet in three-dimensional space, their intersection is not a single point, but a straight line. This fundamental principle is essential in understanding the relationship between surfaces in geometry.

Let $P_1$ and $P_2$ be two intersecting planes. Let their respective vector equations be:

Suppose $\vec{a}$ is the position vector of a point that is common to both planes. Let $\vec{r}$ be the position vector of any other arbitrary point that lies on the intersection of these two planes.

The Geometry of the Intersection Vector

The vector $(\vec{r} - \vec{a})$ represents a line segment that lies entirely on the line of intersection. Therefore, this vector must lie within both planes $P_1$ and $P_2$.

1. Since $(\vec{r} - \vec{a})$ lies in plane $P_1$, it must be perpendicular to the normal vector $\vec{n}_1$.

2. Since $(\vec{r} - \vec{a})$ lies in plane $P_2$, it must be perpendicular to the normal vector $\vec{n}_2$.

Because $(\vec{r} - \vec{a})$ is perpendicular to both $\vec{n}_1$ and $\vec{n}_2$, it must be parallel to their cross product, $\vec{n}_1 \times \vec{n}_2$.

Let $\vec{b} = \vec{n}_1 \times \vec{n}_2$. We can rewrite the expression as:

This confirms that the intersection of two planes is a straight line passing through point $\vec{a}$ and parallel to the vector $(\vec{n}_1 \times \vec{n}_2)$.

Answer:

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Planes Through the Intersection of Two Planes

When two planes intersect, they form a straight line. Any plane that passes through this line of intersection is called a member of the family of planes. Mathematically, any linear combination of the equations of two intersecting planes will result in a new plane that passes through their common line.

Vector Form

Consider two planes, $P_1$ and $P_2$, that are not parallel. Their orientations are defined by their Normal Vectors $\vec{n}_1$ and $\vec{n}_2$. Their standard vector equations are:

Derivation Logic

If we have a point with position vector $\vec{r}$ that lies on the line of intersection of these two planes, this point must simultaneously satisfy the equations of both planes. Mathematically, this implies that for such a vector $\vec{r}$:

We can now construct a linear combination of these two zero-valued expressions. For any real number (scalar parameter) $\lambda$, the following equation must also hold true for every point on the intersection line:

$(\vec{r} \cdot \vec{n}_1 - d_1) + \lambda(\vec{r} \cdot \vec{n}_2 - d_2) = 0$

By applying the distributive property of the dot product, we can group the terms involving $\vec{r}$:

$\vec{r} \cdot \vec{n}_1 + \lambda(\vec{r} \cdot \vec{n}_2) - d_1 - \lambda d_2 = 0$

$\vec{r} \cdot \vec{n}_1 + \vec{r} \cdot (\lambda \vec{n}_2) = d_1 + \lambda d_2$

Cartesian Form

Let the two intersecting planes be defined in Cartesian coordinates as:

Plane 1 ($P_1$): $a_1x + b_1y + c_1z + d_1 = 0$

Plane 2 ($P_2$): $a_2x + b_2y + c_2z + d_2 = 0$

The equation of any plane passing through their line of intersection is given by the linear combination:

Or simply: $P_1 + kP_2 = 0$, where $k$ is a real parameter.

Important Remark:

The equation $P_1 + kP_2 = 0$ can represent every plane in the family except the plane $P_2 = 0$ (since no finite value of $k$ can eliminate $P_1$). If the required plane happens to be $P_2$, we would use the form $P_2 + kP_1 = 0$ instead.

Answer:

Angle Between Two Planes

The inclination between two intersecting planes is mathematically determined by the inclination between their respective normal vectors. As illustrated in geometric principles, the angle $\theta$ between two planes $P_1$ and $P_2$ is defined as the angle between the two lines that are perpendicular (normals) to these planes.

Vector Form

Let the vector equations of the two given planes be:

Here, $\vec{n}_1$ and $\vec{n}_2$ are the vectors normal to plane $P_1$ and plane $P_2$ respectively. If $\theta$ is the angle between these planes, it corresponds exactly to the angle between the vectors $\vec{n}_1$ and $\vec{n}_2$. Using the definition of the dot product, we have:

Cartesian Form

Let the Cartesian equations of the two planes be expressed as:

$a_1x + b_1y + c_1z + d_1 = 0$

$a_2x + b_2y + c_2z + d_2 = 0$

The direction ratios (also known as attitude numbers) of the normals to these planes are $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ respectively. Thus, the normal vectors can be written as:

$\vec{n}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$

$\vec{n}_2 = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$

Substituting these components into the cosine formula derived in the vector form equation:

Acute Angle Calculation

In practice, the intersection of two planes creates two supplementary angles (one acute and one obtuse). By convention, the acute angle is usually required. To ensure the result is acute, we take the absolute value of the expression:

Answer:

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Special Relationships and Conditions Between Two Planes

In Three-Dimensional Geometry, the relative orientation of two planes is dictated by the behavior of their normal vectors. By analyzing these vectors, we can derive the conditions for planes to be perpendicular or parallel, and also determine the general equation for a family of parallel planes.

Condition of Perpendicularity

Two planes are considered perpendicular if and only if the angle between them is $90^\circ$. Since the angle between planes is defined as the angle between their normals, the normal vectors must also be perpendicular.

Vector Form

Let the normal vectors to the two planes be $\vec{n}_1$ and $\vec{n}_2$. The planes are perpendicular iff:

Cartesian Form

If the planes are $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$, their direction ratios are $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$. Substituting these into the dot product:

Condition for Parallelism

Two planes are parallel if and only if their normal vectors are parallel to each other. This means one normal vector is a scalar multiple of the other.

Vector Form

The planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ are parallel iff:

Cartesian Form

In Cartesian coordinates, this implies that the direction ratios (attitude numbers) of the normals must be proportional:

Planes Parallel to a Given Plane

Any plane that is parallel to a given plane will share the same normal direction. Therefore, the coefficients of $x, y,$ and $z$ remain the same (or proportional), while only the constant term differs.

Vector Form

Given a plane with the equation $\vec{r} \cdot \vec{n} = d$, the equation of any plane parallel to it is:

In this form, $\lambda$ is an arbitrary real number that determines the distance of the new plane from the origin.

Cartesian Form

Given a plane $ax + by + cz + d = 0$, any plane parallel to it is represented as:

Here, $k$ is an arbitrary constant. Note that the attitude numbers $\langle a, b, c \rangle$ remain identical because the orientation of the plane has not changed.

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Angle Between a Line and a Plane

The angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane. Intuitively, it is the angle formed between the line and its projection on the plane.

Vector Form

Let the given plane be $P$ and the line be $L$. Their vector equations are:

From the geometric figure, it is evident that if $\theta$ is the angle between the line $L$ and the plane $P$, then the angle between the line $L$ and the normal vector $\vec{n}$ to the plane is $(90^\circ - \theta)$.

Derivation

Using the definition of the dot product between the direction vector of the line ($\vec{b}$) and the normal vector of the plane ($\vec{n}$):

Since $\cos(90^\circ - \theta) = \sin \theta$, we get:

Remark: Since we usually consider the acute angle, we take the absolute value of the dot product. As $0 \leq \theta \leq \frac{\pi}{2}$, $\sin \theta$ remains non-negative.

Cartesian Form

Let the equation of the plane be $Ax + By + Cz + D = 0$ and the equation of the line be $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.

Here:

1. The direction ratios of the line are $\langle a, b, c \rangle$, so $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

2. The attitude numbers (direction ratios of the normal) of the plane are $\langle A, B, C \rangle$, so $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

Final Formula

Substituting these into the vector formula from the equation provided above:

Answer:

Distance of a Point from a Plane

The distance of a point from a plane is defined as the length of the perpendicular segment from the point to the plane. In vector geometry, this is a fundamental concept used to find the shortest distance between a point and a flat surface in three-dimensional space.

Vector form

The length $p$ of the perpendicular drawn from the point $P$ with position vector $\vec{a}$ to the plane $\vec{r} \cdot \vec{n} = d$ is given by:

If the equation of the plane is given in the form $\vec{r} \cdot \vec{n} + d = 0$, the formula is adjusted to $p = \frac{|\vec{a} \cdot \vec{n} + d|}{|\vec{n}|}$.

Derivation of the Formula

Given:

A point $P$ with position vector $\vec{a}$ and a plane with vector equation:

Here, $\vec{n}$ is the vector normal to the plane.

To Find:

The length of the perpendicular segment $p$ from point $P$ to the plane.

Construction Required:

Let $M$ be the foot of the perpendicular drawn from point $P(\vec{a})$ to the plane provided above. The line segment $MP$ represents the perpendicular distance. Since $MP$ is perpendicular to the plane, it is parallel to the normal vector $\vec{n}$.

Proof:

As the line $MP$ passes through the point $P(\vec{a})$ and is parallel to the vector $\vec{n}$, the equation of line $MP$ is:

The point $M$ lies on the line $MP$, so its position vector is of the form $\vec{a} + \lambda \vec{n}$ for some value of $\lambda$. Since $M$ also lies on the plane provided above, we substitute this into the plane equation:

Solving for $\lambda$:

The vector $\vec{MP}$ is calculated as the difference between the position vectors of $P$ and $M$:

The length $p$ is the magnitude of the vector $\vec{MP}$:

Substituting the value of $\lambda$ from equation (i):

This magnitude ensures the distance is always positive, thus:

Distance from the Origin

To find the distance $p$ from the origin $O$ to the plane $\vec{r} \cdot \vec{n} = d$, we substitute the position vector of the origin $\vec{a} = \vec{0}$ into the general distance formula.

In the special case where the plane is given in the normal form $\vec{r} \cdot \hat{n} = d$, where $\hat{n}$ is a unit vector, the distance is simply $|d|$ as $|\hat{n}| = 1$.

Foot of the Perpendicular

The position vector of the foot of the perpendicular $M$ drawn from point $P(\vec{a})$ to the plane $\vec{r} \cdot \vec{n} = d$ is given by the expression:

Where the scalar $\lambda$ is defined as:

This allows for the determination of the exact coordinates of the point on the plane surface which is closest to the given point $P$.

Answer:

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Cartesian Form

The length $p$ of the perpendicular drawn from a point $P(x_1, y_1, z_1)$ to a plane defined by the general Cartesian equation $Ax + By + Cz + D = 0$ represents the shortest distance between the point and the plane.

The formula for this distance is given by:

Derivation of the Cartesian Formula

Given:

1. A point $P(x_1, y_1, z_1)$.

2. A plane equation in Cartesian form:

Construction Required:

Let $M$ be the foot of the perpendicular drawn from $P(x_1, y_1, z_1)$ to the plane provided above. Let the coordinates of $M$ be $(x, y, z)$. The line segment $MP$ is perpendicular to the plane, meaning its direction is parallel to the normal vector of the plane.

Proof:

The direction numbers of the normal to the plane provided above are $\langle A, B, C \rangle$. Since $MP$ is parallel to the normal, the equations of the line $MP$ passing through $P(x_1, y_1, z_1)$ are:

Any general point on the line $MP$ can be written as $(x_1 + A\lambda, y_1 + B\lambda, z_1 + C\lambda)$. This point will coincide with the foot of the perpendicular $M$ if and only if it lies on the plane provided above. Substituting these coordinates into equation the equation of the plane provided above:

Solving for $\lambda$:

The distance $MP$ is the length of the segment between $(x_1, y_1, z_1)$ and $(x_1 + A\lambda, y_1 + B\lambda, z_1 + C\lambda)$. Using the distance formula:

Substituting the value of $\lambda$ from (i):

Distance between Parallel Planes

To find the perpendicular distance between two parallel planes, say $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$:

1. Take any arbitrary point $P(x_1, y_1, z_1)$ on the first plane.

2. Calculate the perpendicular distance from this point $P$ to the second plane using the distance formula derived above.

Note: An alternate direct formula for the distance $d$ between parallel planes is:

Answer: