Shortest Distance Between Skew Lines: Formula & Derivation

Concept Overview

This question tests the understanding of three-dimensional geometry, specifically the concept of skew lines and how to calculate the shortest distance between them. Skew lines are lines in 3D space that are neither parallel nor intersecting. The shortest distance between them is along a line segment perpendicular to both. We will derive the formula for this distance using vector algebra, leveraging the properties of the cross product.

Step 1: Define the equations of the two skew lines. Let the equations of the two skew lines, $L_1$ and $L_2$, in vector form be: $\mathbf{r}_1 = \mathbf{a}_1 + \lambda \mathbf{b}_1$ $\mathbf{r}_2 = \mathbf{a}_2 + \mu \mathbf{b}_2$ Here, $\mathbf{a}_1$ and $\mathbf{a}_2$ are position vectors of points on $L_1$ and $L_2$ respectively, and $\mathbf{b}_1$ and $\mathbf{b}_2$ are direction vectors of $L_1$ and $L_2$. $\lambda$ and $\mu$ are scalar parameters.

Step 2: Understand the geometric interpretation of the shortest distance. The shortest distance between two skew lines is the length of the line segment that is perpendicular to both lines. Let this shortest distance be $d$. This segment lies along a vector that is perpendicular to both direction vectors $\mathbf{b}_1$ and $\mathbf{b}_2$.

Step 3: Identify a vector perpendicular to both direction vectors. A vector perpendicular to both $\mathbf{b}_1$ and $\mathbf{b}_2$ can be found using their cross product: $\mathbf{b}_1 \times \mathbf{b}_2$. This resulting vector is normal to the plane containing lines parallel to $L_1$ and $L_2$ passing through the origin.

Step 4: Consider the vector connecting a point on each line. Let $P_1$ be a point on $L_1$ with position vector $\mathbf{a}_1$, and $P_2$ be a point on $L_2$ with position vector $\mathbf{a}_2$. The vector connecting these two points is $\mathbf{a}_2 - \mathbf{a}_1$.

Step 5: Project the connecting vector onto the common perpendicular. The shortest distance $d$ is the scalar projection of the vector $\mathbf{a}_2 - \mathbf{a}_1$ onto the direction of the common perpendicular vector $\mathbf{b}_1 \times \mathbf{b}_2$. The scalar projection of a vector $\mathbf{v}$ onto a vector $\mathbf{u}$ is given by $\frac{|\mathbf{v} \cdot \mathbf{u}|}{|\mathbf{u}|}$.

Step 6: Apply the scalar projection formula. Using the formula for scalar projection, the shortest distance $d$ is: $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ The term $(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)$ is the scalar triple product, which represents the volume of the parallelepiped formed by the vectors $\mathbf{a}_2 - \mathbf{a}_1$, $\mathbf{b}_1$, and $\mathbf{b}_2$. The magnitude of the cross product $|\mathbf{b}_1 \times \mathbf{b}_2|$ represents the area of the parallelogram formed by $\mathbf{b}_1$ and $\mathbf{b}_2$. The ratio gives the height of the parallelepiped with respect to this base, which is precisely the shortest distance between the skew lines.

Step 7: Alternative form using determinant. The scalar triple product can also be expressed as a determinant: $(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) = \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix}$ where $\mathbf{a}_1 = (x_1, y_1, z_1)$, $\mathbf{a}_2 = (x_2, y_2, z_2)$, $\mathbf{b}_1 = (b_{1x}, b_{1y}, b_{1z})$, and $\mathbf{b}_2 = (b_{2x}, b_{2y}, b_{2z})$. Thus, the formula for the shortest distance becomes: $d = \frac{1}{|\mathbf{b}_1 \times \mathbf{b}_2|} \left| \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix} \right|$

Key Takeaways:

• Skew lines are lines in 3D that are neither parallel nor intersecting.
• The shortest distance between skew lines is the length of the common perpendicular segment.
• The direction of the common perpendicular is given by the cross product of the direction vectors of the lines ($\mathbf{b}_1 \times \mathbf{b}_2$).
• The shortest distance is the scalar projection of the vector connecting any point on one line to any point on the other line, onto the common perpendicular direction.

Answer: $d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$