Three Dimensional Geometry: Distances

Shortest Distance between Two Skew Lines

In three-dimensional space, two straight lines can be intersecting, parallel, or skew. We have already discussed intersecting and parallel lines. Now, we consider the case of skew lines, which are lines that are neither parallel nor intersecting.

Definition of Skew Lines

Two lines in space are called skew lines if and only if they satisfy two conditions simultaneously:

1. They are not parallel.
2. They do not intersect.

Skew lines cannot lie in the same plane. If two lines lie in the same plane, they must either intersect or be parallel. Conversely, if two lines are skew, they must lie in different planes.

Concept of Shortest Distance between Skew Lines

For any two skew lines, there exists a unique line segment that is perpendicular to both lines. The length of this unique segment is defined as the shortest distance between the two skew lines. Imagine holding two pencils in 3D space such that they are not parallel and do not touch; the shortest distance is the length of a rubber band stretched tightly between them, perpendicular to both pencils.

Formula for Shortest Distance using Vector Form

Let the two skew lines $L_1$ and $L_2$ be given by their vector equations:

Here:

• $\vec{a_1}$ is the position vector of a point A that lies on line $L_1$.
• $\vec{b_1}$ is a vector parallel to line $L_1$ (its direction vector). $\vec{b_1} \neq \vec{0}$.
• $\vec{a_2}$ is the position vector of a point B that lies on line $L_2$.
• $\vec{b_2}$ is a vector parallel to line $L_2$ (its direction vector). $\vec{b_2} \neq \vec{0}$.

Since the lines are skew, their direction vectors are not parallel, meaning $\vec{b_1}$ is not a scalar multiple of $\vec{b_2}$. Also, the lines do not intersect.

Let PQ be the unique shortest distance line segment between $L_1$ and $L_2$, where P is on $L_1$ and Q is on $L_2$. This segment PQ is perpendicular to both line $L_1$ and line $L_2$. Therefore, the vector $\vec{PQ}$ is perpendicular to both the direction vector of $L_1$ ($\vec{b_1}$) and the direction vector of $L_2$ ($\vec{b_2}$).

A vector that is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$ is given by their cross product, $\vec{b_1} \times \vec{b_2}$. Since $L_1$ and $L_2$ are not parallel, $\vec{b_1}$ and $\vec{b_2}$ are not collinear, so $\vec{b_1} \times \vec{b_2} \neq \vec{0}$.

The direction of the shortest distance vector $\vec{PQ}$ is along the direction of $\vec{b_1} \times \vec{b_2}$. Let $\hat{n}$ be the unit vector in the direction of $\vec{b_1} \times \vec{b_2}$.

Consider the vector $\vec{AB}$ joining any point A on $L_1$ to any point B on $L_2$. This vector is $\vec{AB} = \vec{a_2} - \vec{a_1}$. The shortest distance $d$ between the lines is the length of the projection of the vector $\vec{AB}$ onto the direction of the common perpendicular, which is $\hat{n}$. This is the absolute value of the scalar projection of $\vec{AB}$ onto $\hat{n}$.

Using the formula for scalar projection of $\vec{u}$ onto $\vec{v}$ as $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$ or $\vec{u} \cdot \hat{v}$:

Substitute the expressions for $\vec{AB}$ and $\hat{n}$:

Using properties of scalar multiplication and the absolute value:

The numerator $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is the scalar triple product of the vectors $\vec{a_2} - \vec{a_1}$, $\vec{b_1}$, and $\vec{b_2}$. It can be written in bracket notation as $[\vec{a_2}-\vec{a_1} \,\, \vec{b_1} \,\, \vec{b_2}]$.

So, the formula for the shortest distance $d$ between two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$ is:

$$ \mathbf{d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}} $$

Condition for Intersecting Lines: The two lines $L_1$ and $L_2$ will intersect if and only if the shortest distance between them is zero. This occurs when the numerator of the formula is zero, i.e., $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$. This means the scalar triple product of $\vec{a_2} - \vec{a_1}$, $\vec{b_1}$, and $\vec{b_2}$ is zero. Geometrically, this means the vectors $\vec{a_2} - \vec{a_1}$, $\vec{b_1}$, and $\vec{b_2}$ are coplanar. The vector $\vec{a_2} - \vec{a_1}$ joins a point on $L_1$ to a point on $L_2$. If this vector is coplanar with the direction vectors of the two lines, it means the two lines lie in the same plane, and since they are not parallel (as they are skew or intersecting), they must intersect.

Formula for Shortest Distance using Cartesian Form

Let the Cartesian equations of the two skew lines $L_1$ and $L_2$ be:

From these equations, we can identify:

• A point on $L_1$ is A$(x_1, y_1, z_1)$. Its position vector is $\vec{a_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.
• The direction ratios of $L_1$ are $(a_1, b_1, c_1)$. A direction vector is $\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$.
• A point on $L_2$ is B$(x_2, y_2, z_2)$. Its position vector is $\vec{a_2} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$.
• The direction ratios of $L_2$ are $(a_2, b_2, c_2)$. A direction vector is $\vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$.

The vector connecting the two points is $\vec{a_2} - \vec{a_1} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.

The numerator of the shortest distance formula is the absolute value of the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$. In component form, this is the absolute value of the determinant:

The denominator is the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$. First, calculate the cross product in component form:

Then its magnitude is:

Combining these, the shortest distance $d$ between two skew lines in Cartesian form is:

$$ \mathbf{d = \frac{\left| \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \right|}{\sqrt{(b_1c_2 - b_2c_1)^2 + (a_1c_2 - a_2c_1)^2 + (a_1b_2 - a_2b_1)^2}}} $$

Answer:

Shortest Distance between Parallel Lines

Parallel lines, unlike skew lines, lie in the same plane. This simplifies the problem of finding the shortest distance between them. The shortest distance between two parallel lines is the constant perpendicular distance between any point on one line and the other line.

Concept

Let the two parallel lines $L_1$ and $L_2$ be given by their vector equations. Since they are parallel, they must have direction vectors that are collinear. We can represent them using the same direction vector $\vec{b}$.

Here, $\vec{a_1}$ is the position vector of a point A on $L_1$, and $\vec{a_2}$ is the position vector of a point B on $L_2$. $\vec{b}$ is the common direction vector of the parallel lines.

The shortest distance between $L_1$ and $L_2$ is the length of the perpendicular segment from any point on one line to the other line. Let's find the perpendicular distance from point B (on $L_2$) to line $L_1$.

Formula for Shortest Distance using Vector Form

Consider the vector $\vec{AB}$ joining point A on $L_1$ to point B on $L_2$. $\vec{AB} = \vec{a_2} - \vec{a_1}$.

Let $\theta$ be the angle between the vector $\vec{AB}$ and the direction vector $\vec{b}$ (where $0 \le \theta \le \pi$).

Imagine a parallelogram formed by the vectors $\vec{AB}$ and $\vec{b}$ (if we were to place $\vec{b}$ at A). The area of this parallelogram is $|\vec{AB} \times \vec{b}|$. This area is also equal to the base times the height. Let $|\vec{b}|$ be the base. The height $h$ of this parallelogram relative to the base $|\vec{b}|$ is the perpendicular distance from B to the line $L_1$. This height is the shortest distance $d$ we are looking for.

Area of parallelogram $= |\vec{b}| \times d$

Also, Area of parallelogram $= |\vec{AB} \times \vec{b}|$.

Equating the two expressions for the area:

Assuming $\vec{b} \neq \vec{0}$ (which is true for a line's direction vector), we can solve for $d$:

Substitute $\vec{AB} = \vec{a_2} - \vec{a_1}$:

$$ \mathbf{d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}} $$

This formula gives the shortest distance $d$ between two parallel lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b}$.

Note: This formula is also valid if the lines are coincident (the same line). In that case, point B lies on $L_1$, so $\vec{a_2} - \vec{a_1}$ is parallel to $\vec{b}$. The cross product $(\vec{a_2} - \vec{a_1}) \times \vec{b}$ would be $\vec{0}$, and the distance $d$ would be $\frac{|\vec{0}|}{|\vec{b}|} = 0$, which is correct for coincident lines.

Formula using Cartesian Form

If the Cartesian equations of the two parallel lines $L_1$ and $L_2$ are given:

Notice that the direction ratios $(a, b, c)$ are the same for both lines, indicating they are parallel. We can identify:

• A point on $L_1$: A$(x_1, y_1, z_1)$. Position vector $\vec{a_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.
• A point on $L_2$: B$(x_2, y_2, z_2)$. Position vector $\vec{a_2} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$.
• The common direction vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

The vector connecting the points is $\vec{a_2} - \vec{a_1} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.

The shortest distance formula in Cartesian form is obtained by substituting these into the vector formula $d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}$.

$$ \mathbf{d = \frac{|((x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}) \times (a\hat{i} + b\hat{j} + c\hat{k})|}{\sqrt{a^2 + b^2 + c^2}}} $$

The numerator requires calculating the magnitude of a cross product. The cross product is:

The magnitude of this cross product is $\sqrt{((y_2-y_1)c - (z_2-z_1)b)^2 + ((x_2-x_1)c - (z_2-z_1)a)^2 + ((x_2-x_1)b - (y_2-y_1)a)^2}$.

So, the shortest distance is this magnitude divided by $\sqrt{a^2 + b^2 + c^2}$. While the formula looks complicated in full Cartesian form, it's easier to compute by finding the vectors $\vec{a_2}-\vec{a_1}$ and $\vec{b}$ and using the vector formula directly.

Answer:

Distance of a Point from a Plane

One common problem in three-dimensional geometry is finding the shortest distance from a given point in space to a given plane. The shortest distance is always measured along the line segment that is perpendicular to the plane and connects the point to the plane.

Formula for Distance of a Point from a Plane using Vector Form

Let P be the given point with position vector $\vec{p}$ with respect to the origin O.

Let the equation of the given plane be in the form $\vec{r} \cdot \vec{N} = d$. Here, $\vec{N}$ is a vector normal (perpendicular) to the plane, and $d$ is a scalar constant. (This form is equivalent to $\vec{r} \cdot \vec{N} - d = 0$).

Let A be any point that lies on the plane. Let its position vector be $\vec{a} = \vec{OA}$. Since A lies on the plane, its position vector must satisfy the plane's equation:

Consider the vector $\vec{AP}$ joining the point A (on the plane) to the given point P (outside the plane). This vector is $\vec{AP} = \vec{OP} - \vec{OA} = \vec{p} - \vec{a}$.

The shortest distance $D$ from point P to the plane is the length of the perpendicular segment from P to the plane. This length is equal to the magnitude of the scalar projection of the vector $\vec{AP}$ onto the normal vector $\vec{N}$.

The unit vector in the direction of the normal vector $\vec{N}$ is $\hat{n} = \frac{\vec{N}}{|\vec{N}|}$.

The scalar projection of $\vec{AP}$ onto $\vec{N}$ is $\vec{AP} \cdot \hat{n}$. The distance $D$ is the absolute value of this scalar projection:

Using properties of scalar multiplication and absolute value:

Use the distributive property of the dot product in the numerator:

Since A is a point on the plane, we know $\vec{a} \cdot \vec{N} = d$. Substitute this into the numerator:

$$ \mathbf{D = \frac{|\vec{p} \cdot \vec{N} - d|}{|\vec{N}|}} $$

This formula gives the perpendicular distance from the point with position vector $\vec{p}$ to the plane with vector equation $\vec{r} \cdot \vec{N} = d$.

If the plane equation is given in the normal form $\vec{r} \cdot \hat{n} = p$, where $\hat{n}$ is a unit normal vector and $p$ is the perpendicular distance from the origin, then $\vec{N} = \hat{n}$ and $|\vec{N}|=|\hat{n}|=1$, and $d=p$. The distance formula simplifies to:

$$ \mathbf{D = \frac{|\vec{p} \cdot \hat{n} - p|}{1} = |\vec{p} \cdot \hat{n} - p|} $$

Formula for Distance of a Point from a Plane using Cartesian Form

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

Let the equation of the given plane be in the general Cartesian form $Ax + By + Cz + D = 0$.

The normal vector to the plane is $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$. Its magnitude is $|\vec{N}| = \sqrt{A^2 + B^2 + C^2}$.

The Cartesian equation $Ax + By + Cz + D = 0$ can be written in vector form as $\vec{r} \cdot \vec{N} = -D$. (To see this, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then $\vec{r} \cdot \vec{N} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (A\hat{i} + B\hat{j} + C\hat{k}) = Ax + By + Cz$. So, $Ax + By + Cz = -D$, which is equivalent to $Ax + By + Cz + D = 0$).

Comparing $\vec{r} \cdot \vec{N} = -D$ with the form $\vec{r} \cdot \vec{N} = d$, we have $d = -D$.

Now substitute the position vector $\vec{p} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$, the normal vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$, and $d = -D$ into the vector distance formula $D_{cart} = \frac{|\vec{p} \cdot \vec{N} - d|}{|\vec{N}|}$.

First, calculate $\vec{p} \cdot \vec{N}$ in component form:

Substitute this and the values for $d$ and $|\vec{N}|$ into the distance formula:

$$ \mathbf{D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}} $$

This is the formula for the perpendicular distance from the point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$.

Interpretation: To find the distance, substitute the coordinates of the point $(x_1, y_1, z_1)$ into the left-hand side of the plane's equation $Ax + By + Cz + D$. Take the absolute value of the result. Divide this by the magnitude of the normal vector's coefficients $\sqrt{A^2 + B^2 + C^2}$.

As a check, the distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is $\frac{|A(0) + B(0) + C(0) + D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$, which is consistent with the distance $p$ from the origin derived when converting to the normal form $lx+my+nz=p$.

Answer:

Distance of a Point from a Line

Finding the shortest (perpendicular) distance from a given point to a given line in three-dimensional space is another common distance problem. Unlike the point-plane distance, this is calculated using the cross product.

Concept

Let P be the given point in space, with position vector $\vec{p}$ relative to the origin O.

Let the given line L be represented by the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$. This line passes through a point A with position vector $\vec{a}$ and is parallel to the direction vector $\vec{b}$.

We want to find the length of the perpendicular segment from point P to the line L. Let M be the foot of this perpendicular on the line L. The shortest distance is the length of the segment PM, denoted by $d$.

Formula for Distance of a Point from a Line using Cross Product

Consider the vector $\vec{AP}$ joining the point A (on the line) to the given point P (outside the line). This vector is $\vec{AP} = \vec{OP} - \vec{OA} = \vec{p} - \vec{a}$.

Let $\theta$ be the angle between the vector $\vec{AP}$ and the direction vector of the line $\vec{b}$ (where $0 \le \theta \le \pi$).

In the right-angled triangle $\triangle APM$, the perpendicular distance $d = PM$ is related to the magnitude of $\vec{AP}$ and the angle $\theta$ by trigonometry:

Recall the definition of the magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$: $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \phi$, where $\phi$ is the angle between them. Applying this to vectors $\vec{AP}$ and $\vec{b}$:

We can solve this equation for $|\vec{AP}| \sin \theta$ (assuming $|\vec{b}| \neq 0$):

Since $d = |\vec{AP}| \sin \theta$, we have the formula for the shortest distance $d$ from point P to the line L:

$$ \mathbf{d = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|}} $$

This formula provides the distance using the position vector of the point, the position vector of a point on the line, and the direction vector of the line.

Steps for Calculation using Vector Form

To find the distance of a point P($\vec{p}$) from a line $\vec{r} = \vec{a} + \lambda \vec{b}$:

1. Identify the position vector $\vec{p}$ of the given point P. If coordinates $(x_1, y_1, z_1)$ are given, $\vec{p} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.
2. From the line equation $\vec{r} = \vec{a} + \lambda \vec{b}$, identify the position vector $\vec{a}$ of a point A on the line and the direction vector $\vec{b}$ of the line.
3. Calculate the vector joining the point A on the line to the given point P: $\vec{AP} = \vec{p} - \vec{a}$.
4. Calculate the cross product of this joining vector and the line's direction vector: $(\vec{p} - \vec{a}) \times \vec{b}$.
5. Find the magnitude of this cross product: $|(\vec{p} - \vec{a}) \times \vec{b}|$.
6. Find the magnitude of the line's direction vector: $|\vec{b}|$.
7. Divide the magnitude found in step 5 by the magnitude found in step 6 to get the distance $d$.

Cartesian Approach (Implicit)

While there isn't a single compact formula in Cartesian coordinates analogous to the point-plane distance formula, one can derive the distance by converting the vectors in the vector formula into their component forms. If the point is $(x_1, y_1, z_1)$ and the line is $\frac{x - x_2}{a} = \frac{y - y_2}{b} = \frac{z - z_2}{c}$, we would set $\vec{p} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$, $\vec{a} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$, and then follow the steps for the vector formula.

Alternatively, one could find the coordinates of the foot of the perpendicular M on the line L, and then calculate the distance PM using the distance formula between two points. The point M lies on the line, so its coordinates are $(x_2 + \lambda a, y_2 + \lambda b, z_2 + \lambda c)$ for some $\lambda$. The vector $\vec{PM} = \vec{m} - \vec{p} = (x_2 + \lambda a - x_1)\hat{i} + \dots$ must be perpendicular to the direction vector $\vec{b}$. Their dot product $\vec{PM} \cdot \vec{b} = 0$ allows us to solve for $\lambda$. Once $\lambda$ is found, the coordinates of M are known, and the distance PM can be calculated.

Answer: