Question

Find the distance between the line \( \frac{x}{2} = \frac{2y - 6}{4} = \frac{1 - z}{-1} \) 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: \[ \frac{x}{2} = \frac{2y - 6}{4} = \frac{1 - z}{-1} \implies \vec{r}_1 = \vec{0} + \lambda(2\hat{i} + \hat{j} + \hat{k}) \] The line \( L_2 \) parallel to \( L_1 \) and passing 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
Let \( \vec{a}_2 - \vec{a}_1 = (4\hat{i} - 5\hat{k}) - (0) = 4\hat{i} - 5\hat{k} \). The direction vector \( \vec{b} = 2\hat{i} + \hat{j} + \hat{k} \). 
Step 3: Find the shortest distance
The shortest distance \( S.D. \) is given by: \[ {S.D.} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b})|}{|\vec{b}|} \] Compute \( \vec{b} \times (\vec{a}_2 - \vec{a}_1) \): \[ \vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 
2 & 1 & 1 
4 & 0 & -5 \end{vmatrix} = 9\hat{i} - 16\hat{j} + 14\hat{k} \] The magnitude: \[ | \vec{b} | = \sqrt{2^2 + 1^2 + 1^2} = 3 \] \[ {S.D.} = \frac{\sqrt{81 + 256 + 196}}{3} = \frac{\sqrt{533}}{3} \, {units.} \]