Question

Find the distance between the line \[ \frac{x}{2} = \frac{2}{y - 6} = \frac{1}{-z - 1} \] and another line parallel to it passing through the point \( (4, 0, -5) \).

Hint:

For the shortest distance between skew or parallel lines, use the cross-product approach for accuracy.

Answer 1

Step 1: Standardize the equations of the lines
The given line \( L_1 \) is: \[ \frac{x}{2} = \frac{2}{y - 6} = \frac{1}{-z - 1} \] This can be rewritten in vector form as: \[ \vec{r}_1 = \vec{0} + \lambda (2\hat{i} + \hat{j} + \hat{k}) \] The line \( L_2 \), which is parallel to \( L_1 \) and passes through \( (4, 0, -5) \), is: \[ \vec{r}_2 = (4\hat{i} - 5\hat{k}) + \mu (2\hat{i} + \hat{j} + \hat{k}) \]
Step 2: Vector between the lines
The position vector difference between points on the two lines is: \[ \vec{a}_2 - \vec{a}_1 = (4\hat{i} - 5\hat{k}) - (0) = 4\hat{i} - 5\hat{k} \] The direction vector of both lines is: \[ \vec{b} = 2\hat{i} + \hat{j} + \hat{k} \]
Step 3: Find the shortest distance
The shortest distance \( S.D. \) between two skew lines is given by: \[ S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b})|}{|\vec{b}|} \] Compute the cross product \( \vec{b} \times (\vec{a}_2 - \vec{a}_1) \): \[ \vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 4 & 0 & -5 \end{vmatrix} = 9\hat{i} - 16\hat{j} + 14\hat{k} \] The magnitude of \( \vec{b} \) is: \[ |\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \] Thus, the shortest distance is: \[ S.D. = \frac{\sqrt{(81 + 256 + 196)}}{3} = \frac{\sqrt{533}}{3} \text{ units.} \]