Three Dimensional Geometry: Introduction and Lines

Introduction to Three Dimensional Geometry (Coordinate System, Points, Distance Formula, Section Formula - in 3D)

Three-dimensional geometry, often referred to as solid geometry, is the branch of geometry that studies figures and shapes in three-dimensional space. It extends the concepts of two-dimensional coordinate geometry by adding a third dimension. In 3D geometry, we use a coordinate system to locate points and describe lines, planes, surfaces, and solids algebraically.

Rectangular Cartesian Coordinate System in 3D

To locate a point in three-dimensional space, we typically use a rectangular Cartesian coordinate system. This system consists of three mutually perpendicular straight lines that intersect at a single point. These lines are called the coordinate axes, conventionally labeled as the X-axis, the Y-axis, and the Z-axis.

• The point where the three axes intersect is called the origin of the coordinate system. It is usually denoted by the letter O, and its coordinates are $(0, 0, 0)$.
• The axes are oriented such that they form a right-handed system. If you curl the fingers of your right hand from the positive X-axis towards the positive Y-axis, your thumb points in the direction of the positive Z-axis.
• The three axes define three coordinate planes:
  • The XY-plane: This plane contains the X and Y axes. Any point on this plane has its z-coordinate equal to 0. The equation of the XY-plane is $z = 0$.
  • The YZ-plane: This plane contains the Y and Z axes. Any point on this plane has its x-coordinate equal to 0. The equation of the YZ-plane is $x = 0$.
  • The XZ-plane: This plane contains the X and Z axes. Any point on this plane has its y-coordinate equal to 0. The equation of the XZ-plane is $y = 0$.
• These three coordinate planes divide the entire space into eight regions, called octants. Each octant is defined by the signs of the x, y, and z coordinates. For example, the first octant consists of points $(x, y, z)$ where $x > 0$, $y > 0$, and $z > 0$.

Coordinates of a Point in Space

Any point P in three-dimensional space can be uniquely identified by an ordered triplet of real numbers $(x, y, z)$, called the coordinates of the point P. These coordinates represent the perpendicular distances from the point to the three coordinate planes:

• The coordinate $x$ is the perpendicular distance of P from the YZ-plane (signed distance, depending on which side of the plane P lies).
• The coordinate $y$ is the perpendicular distance of P from the XZ-plane.
• The coordinate $z$ is the perpendicular distance of P from the XY-plane.

Conversely, for any ordered triplet $(x, y, z)$ of real numbers, there exists a unique point P in space having these coordinates.

---

Distance Formula in 3D

The distance between any two points in three-dimensional space can be found using an extension of the Pythagorean theorem applied in 3D. Let P be a point with coordinates $(x_1, y_1, z_1)$ and Q be another point with coordinates $(x_2, y_2, z_2)$. The distance between P and Q, denoted as PQ, is given by the formula:

$$ \mathbf{PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}} $$

This formula is a direct consequence of applying the Pythagorean theorem in 3D space. Consider the rectangular box formed by planes through P and Q parallel to the coordinate planes. The distance PQ is the length of the diagonal of this box, and the edge lengths of the box are $|x_2 - x_1|$, $|y_2 - y_1|$, and $|z_2 - z_1|$.

A special case is finding the distance of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$. Using the distance formula with $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (x, y, z)$:

Answer:

---

Section Formula in 3D

The section formula is used to find the coordinates of a point that divides the line segment joining two given points in a specific ratio. This formula directly extends from the 2D version and the vector section formula by applying the ratio to each coordinate separately.

Let P and Q be two points with coordinates P$(x_1, y_1, z_1)$ and Q$(x_2, y_2, z_2)$. Let R be a point with coordinates R$(x, y, z)$ that divides the line segment PQ in the ratio $m:n$.

1. Internal Division

If the point R divides the line segment PQ internally in the ratio $m:n$ (meaning R is between P and Q), its coordinates $(x, y, z)$ are given by:

So, the coordinates of R are $\left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n} \right)$. This corresponds exactly to the component form of the vector section formula $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$ where $\vec{a}=(x_1, y_1, z_1)$, $\vec{b}=(x_2, y_2, z_2)$, and $\vec{r}=(x, y, z)$.

2. External Division

If the point R divides the line segment PQ externally in the ratio $m:n$ (meaning R lies on the line extending PQ, outside the segment, with $m \neq n$), its coordinates $(x, y, z)$ are given by:

So, the coordinates of R are $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n} \right)$. This matches the component form of the vector external section formula $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$.

3. Midpoint Formula

The midpoint M of the line segment PQ is a special case of internal division where the ratio is $1:1$ (i.e., $m=1, n=1$). Substituting these values into the internal division formula gives the coordinates of the midpoint M$(x, y, z)$:

So, the coordinates of the midpoint M are $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$. This is the component form of the vector midpoint formula $\vec{m} = \frac{\vec{a} + \vec{b}}{2}$.

---

Answer:

Direction Cosines and Direction Ratios of a Line

In three-dimensional geometry, specifying the location of points is not enough to describe lines and planes. We also need concepts to describe their orientation in space. For a line, this orientation is given by its direction cosines or direction ratios.

Direction Angles

Consider a directed line L in space. The direction of this line can be described by the angles it makes with the positive directions of the coordinate axes (X, Y, and Z). If the line passes through the origin, these angles are measured directly. If the line does not pass through the origin, we consider a line parallel to L that passes through the origin.

Let the directed line L make angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the X-axis, Y-axis, and Z-axis respectively. These angles are called the direction angles of the directed line L.

By convention, the direction angles are usually taken in the range $0 \le \alpha \le \pi$, $0 \le \beta \le \pi$, $0 \le \gamma \le \pi$ (or $0^\circ$ to $180^\circ$).

A line in space has two opposite directions. If the direction angles for one direction are $\alpha, \beta, \gamma$, then the direction angles for the opposite direction are $\pi - \alpha, \pi - \beta, \pi - \gamma$.

Direction Cosines (DCs)

The direction cosines of a directed line are the cosines of its direction angles.

• They are commonly denoted by the letters $l, m, n$.
• $l = \cos \alpha$
• $m = \cos \beta$
• $n = \cos \gamma$

Consider a point P$(x, y, z)$ on the directed line L passing through the origin, at a distance $r$ from the origin, i.e., $OP = r = \sqrt{x^2+y^2+z^2}$. The coordinates of P can be related to the distance $r$ and the direction cosines:

By projecting the line segment OP onto the axes, we get:

Substitute these into the distance formula from the origin, $x^2 + y^2 + z^2 = r^2$:

Assuming $r \neq 0$ (i.e., P is not the origin), we can divide by $r^2$ to obtain the fundamental identity relating direction cosines:

$$ \mathbf{l^2 + m^2 + n^2 = 1} $$

or equivalently, using the definitions of $l, m, n$:

$$ \mathbf{\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1} $$

This property is crucial: the sum of the squares of the direction cosines of any line is always equal to 1.

If $(l, m, n)$ are the direction cosines for one direction of a line, then the direction cosines for the opposite direction are $(\cos(\pi-\alpha), \cos(\pi-\beta), \cos(\pi-\gamma)) = (-\cos\alpha, -\cos\beta, -\cos\gamma) = (-l, -m, -n)$. Both $(l, m, n)$ and $(-l, -m, -n)$ are valid sets of direction cosines for the same line (undirected).

Direction Ratios (DRs)

Direction ratios, or direction numbers, of a line are any set of three numbers that are proportional to the direction cosines of the line. If $(l, m, n)$ are the direction cosines of a line, then any set of three numbers $(a, b, c)$ such that:

for some non-zero scalar $k$, are called direction ratios of the line.

Alternatively, $(a, b, c)$ are direction ratios if $\frac{l}{a} = \frac{m}{b} = \frac{n}{c}$ for some non-zero constant (assuming $a,b,c \neq 0$).

Unlike direction cosines, direction ratios are not unique. If $(a, b, c)$ are direction ratios, then $(2a, 2b, 2c)$, $(-a, -b, -c)$, etc., are also direction ratios for the same line.

Relationship between Direction Cosines and Direction Ratios

If $(a, b, c)$ are direction ratios of a line, we can find its direction cosines $(l, m, n)$. We know that $l=ka, m=kb, n=kc$ for some constant $k$. From the identity $l^2 + m^2 + n^2 = 1$, we substitute these expressions:

Assuming $a, b, c$ are not all zero, $a^2 + b^2 + c^2 \neq 0$. We can solve for $k^2$:

Taking the square root:

Now substitute this value of $k$ back into the expressions for $l, m, n$:

$$ \mathbf{l = ka = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}} $$

$$ \mathbf{m = kb = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}} $$

$$ \mathbf{n = kc = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}} $$

The sign (either $+$ for all three or $-$ for all three) determines the specific direction of the line. If only the undirected line is considered, either set of direction cosines is valid.

Direction Ratios and Direction Cosines of a Line Joining Two Points

If a line passes through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, we can easily find its direction ratios and direction cosines. Consider the vector $\vec{PQ}$ starting from P and ending at Q. The direction of the line PQ is the same as the direction of the vector $\vec{PQ}$.

The components of the vector $\vec{PQ}$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. These components are proportional to the direction cosines of the line PQ. Therefore, the direction ratios $(a, b, c)$ of the line PQ are given by the differences in the corresponding coordinates:

$$ \mathbf{a = x_2 - x_1} $$

$$ \mathbf{b = y_2 - y_1} $$

$$ \mathbf{c = z_2 - z_1} $$

To find the direction cosines $(l, m, n)$ of the directed line segment $\vec{PQ}$, we divide the direction ratios by the magnitude of the vector $\vec{PQ}$, which is the distance between P and Q, $|\vec{PQ}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

$$ l = \frac{x_2 - x_1}{|\vec{PQ}|} = \frac{x_2 - x_1}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}} $$

$$ m = \frac{y_2 - y_1}{|\vec{PQ}|} = \frac{y_2 - y_1}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}} $$

$$ n = \frac{z_2 - z_1}{|\vec{PQ}|} = \frac{z_2 - z_1}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}} $$

If we consider the line PQ as undirected, the direction cosines could also be the negatives of these values.

---

Answer:

Equation of a Straight Line in Space (Vector Form)

Just as in 2D geometry, a straight line in three-dimensional space is uniquely determined by knowing either one point on the line and its direction, or two distinct points that lie on the line. We can use vector algebra to represent the equation of such a line.

1. Vector Equation of a Line Through a Given Point and Parallel to a Given Vector

Suppose we want to find the equation of a straight line that passes through a specific point A in space and is parallel to a given non-zero vector $\vec{b}$.

Let O be the origin. Let the position vector of the given point A be $\vec{a} = \vec{OA}$.

Let $\vec{b}$ be the given vector parallel to the line. This vector $\vec{b}$ determines the direction of the line and is often called the direction vector of the line.

Let P be any arbitrary point on the line. Let its position vector relative to the origin O be $\vec{r} = \vec{OP}$.

Consider the vector $\vec{AP}$. Since the point P lies on the line that passes through A and is parallel to $\vec{b}$, the vector $\vec{AP}$ must be parallel to the vector $\vec{b}$.

According to the definition of collinear vectors, if $\vec{AP}$ is parallel to $\vec{b}$, then $\vec{AP}$ must be a scalar multiple of $\vec{b}$.

Now, express the vector $\vec{AP}$ in terms of the position vectors of its endpoints A and P:

Substitute this expression for $\vec{AP}$ into the collinearity condition:

Rearranging this equation to solve for $\vec{r}$, which is the position vector of any point on the line, we get the vector equation of the line:

$$ \mathbf{\vec{r} = \vec{a} + \lambda \vec{b}} $$

This is the vector equation of a straight line passing through a point with position vector $\vec{a}$ and parallel to the vector $\vec{b}$.

• $\vec{r}$ represents the position vector of any general point on the line.
• $\vec{a}$ is the position vector of a known fixed point that the line passes through.
• $\vec{b}$ is any vector parallel to the line (the direction vector).
• $\lambda$ is a scalar parameter. As $\lambda$ takes on all possible real values (from $-\infty$ to $+\infty$), the point P with position vector $\vec{r}$ traces out the entire straight line.

2. Vector Equation of a Line Through Two Given Points

Alternatively, a line can be determined by two distinct points that lie on it. Suppose we want to find the equation of a straight line that passes through two distinct points A and B in space.

Let O be the origin. Let the position vectors of the given points A and B be $\vec{a} = \vec{OA}$ and $\vec{b} = \vec{OB}$ respectively.

Let P be any arbitrary point on the line passing through A and B, with position vector $\vec{r} = \vec{OP}$.

Since the points A, B, and P are collinear (they all lie on the same line), the vector $\vec{AP}$ must be collinear with the vector $\vec{AB}$.

Therefore, $\vec{AP}$ must be a scalar multiple of $\vec{AB}$:

Now, express the vectors $\vec{AP}$ and $\vec{AB}$ in terms of position vectors:

Substitute these expressions into the collinearity condition:

Rearranging this equation to solve for $\vec{r}$, we get the vector equation of the line:

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

This is the vector equation of a straight line passing through two points with position vectors $\vec{a}$ and $\vec{b}$.

In this form, $\vec{a}$ is the position vector of one point on the line (point A), and the vector $(\vec{b} - \vec{a})$ is a vector along the line (from A to B), thus serving as the direction vector parallel to the line. This second form is essentially a specific case of the first form, where the direction vector $\vec{b}$ (from the first form) is replaced by the vector joining the two points, $\vec{b} - \vec{a}$.

Alternatively, the equation can be written as $\vec{r} = (1-\lambda)\vec{a} + \lambda\vec{b}$. This form shows that any point on the line is a linear combination of the position vectors of the two points. For $0 \le \lambda \le 1$, $\vec{r}$ represents points on the line segment AB.

Equation of a Straight Line in Space (Cartesian Form)

While vector form is concise and useful for theoretical purposes and certain operations, the equation of a line in space can also be expressed using the Cartesian coordinates $(x, y, z)$ of a general point on the line. This is known as the Cartesian form or symmetric form of the line's equation.

1. Cartesian Equation from a Point and Direction Ratios

Let's start with the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{b}$: $\vec{r} = \vec{a} + \lambda \vec{b}$.

Let the given point be A with coordinates $(x_1, y_1, z_1)$. Its position vector is $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

Let the direction ratios of the line (which are proportional to the components of the parallel vector $\vec{b}$) be $(a, b, c)$. We can take the direction vector $\vec{b}$ to have components equal to the direction ratios, i.e., $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. (Note: If $(l,m,n)$ were the direction cosines, we could use $\vec{b} = l\hat{i} + m\hat{j} + n\hat{k}$).

Let P be any point on the line with coordinates $(x, y, z)$. Its position vector is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

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

Combine the terms on the right side:

Equating the corresponding scalar components of $\hat{i}, \hat{j}, \hat{k}$ on both sides:

These equations are called the parametric equations of the line, with parameter $\lambda$.

If $a, b, c$ are all non-zero, we can solve each equation for $\lambda$:

Since all these expressions are equal to the same parameter $\lambda$, we can set them equal to each other to eliminate $\lambda$. This gives the Cartesian equation (or symmetric form) of the line:

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

This equation represents the line passing through the point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$. Note that the denominators $a, b, c$ are the direction ratios, and the numerators $(x - x_1), (y - y_1), (z - z_1)$ show the differences in coordinates from the fixed point.

Special Cases (when a direction ratio is zero): If one or more direction ratios are zero, the corresponding denominator in the symmetric form is 0. This fraction notation $\frac{x - x_1}{0}$ is interpreted as $x - x_1 = 0$, or $x = x_1$. For example, if the direction ratios are $(a, b, 0)$, the Cartesian equations are $\frac{x - x_1}{a} = \frac{y - y_1}{b}$ and $z - z_1 = 0$ (i.e., $z = z_1$). This indicates that the line lies in the plane $z = z_1$ and its projection onto the XY-plane is given by the 2D line equation $\frac{x - x_1}{a} = \frac{y - y_1}{b}$. If two direction ratios are zero, e.g., $(a, 0, 0)$, the equations are $y = y_1$ and $z = z_1$. This means the line is parallel to the X-axis and passes through $(x_1, y_1, z_1)$.

2. Cartesian Equation of a Line Through Two Given Points

Suppose a line passes through two distinct points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.

The direction ratios $(a, b, c)$ of the line AB are given by the differences in the corresponding coordinates of A and B:

Now, we can use the Cartesian equation form for a line through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$. Using point A$(x_1, y_1, z_1)$ and the calculated direction ratios:

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

This is the Cartesian equation of the line passing through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Note that this form is obtained by substituting the direction ratios $(x_2-x_1, y_2-y_1, z_2-z_1)$ into the point-direction ratio form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The same interpretation regarding zero denominators applies here as well.

---

Answer:

---

Answer:

Conversion between Vector and Cartesian Forms of a Line

We have seen how to represent the equation of a straight line in space using both vector form and Cartesian form. It is often necessary to convert between these two forms, which is a straightforward process as they describe the same geometric object.

1. Converting from Vector Form to Cartesian Form

Given the vector equation of a line:

where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector of any point P$(x, y, z)$ on the line.

$\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is the position vector of a known point A$(x_1, y_1, z_1)$ on the line.

$\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is a vector parallel to the line (the direction vector). Its components $(a, b, c)$ are the direction ratios of the line.

Substitute these into the vector equation:

Combine the terms on the right-hand side by grouping the components:

Equating the corresponding components of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides, we obtain the parametric equations of the line:

Assuming $a, b, c$ are all non-zero, we can solve each equation for the parameter $\lambda$:

Since all these expressions are equal to the same parameter $\lambda$, we can equate them to eliminate $\lambda$ and obtain the Cartesian equation:

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

This formula maps directly from the vector form $\vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda (a\hat{i} + b\hat{j} + c\hat{k})$ to the Cartesian form.

Handling Zero Denominators: If any of $a, b,$ or $c$ is zero, the corresponding term in the symmetric Cartesian form is interpreted as setting the numerator to zero. For example, if $a=0$, the equations are $x = x_1$ and $\frac{y-y_1}{b} = \frac{z-z_1}{c}$.

2. Converting from Cartesian Form to Vector Form

Given the Cartesian equation of a line:

To convert this to vector form, we need to identify a point on the line and the direction ratios of the line.

From the structure of the Cartesian equation:

• The coordinates of a known point on the line are $(x_1, y_1, z_1)$. The position vector of this point is $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.
• The direction ratios of the line are the denominators $(a, b, c)$. A vector parallel to the line (the direction vector) can be taken as $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. (Any non-zero scalar multiple of this vector is also a valid direction vector).

Now, we introduce a scalar parameter, say $\lambda$, by setting each part of the Cartesian equation equal to $\lambda$:

These are the parametric equations of the line. Now, write the position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ using these expressions for $x, y, z$:

Group the terms that do not depend on $\lambda$ and the terms that do:

Factor out $\lambda$ from the second group:

This is the vector form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

Handling Zero Denominators: If the Cartesian equation is given in the form where one or more denominators are zero (e.g., $x = x_1, \frac{y - y_1}{b} = \frac{z - z_1}{c}$), we still identify the point $(x_1, y_1, z_1)$. The zero denominators indicate zero direction ratios in those components. So, if the equations are $x=x_1, y=y_1, \frac{z-z_1}{c}$, the direction ratios are $(0, 0, c)$. The direction vector is $c\hat{k}$ (or just $\hat{k}$). The vector equation is $\vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda c\hat{k}$.

---

Answer:

Answer: