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 equation for line \( L_1 \) is:\[\frac{x}{2} = \frac{2y - 6}{4} = \frac{1 - z}{-1} \implies \vec{r}_1 = \vec{0} + \lambda(2\hat{i} + \hat{j} + \hat{k})\]Line \( L_2 \) is parallel to \( L_1 \) and passes through \( (4, 0, -5) \). Its equation is:\[\vec{r}_2 = (4\hat{i} - 5\hat{k}) + \mu(2\hat{i} + \hat{j} + \hat{k})\]Step 2: Determine the vector connecting the lines
The vector connecting a point on \( L_2 \) to a point on \( L_1 \) is \( \vec{a}_2 - \vec{a}_1 = (4\hat{i} - 5\hat{k}) - (0) = 4\hat{i} - 5\hat{k} \). The direction vector is \( \vec{b} = 2\hat{i} + \hat{j} + \hat{k} \).Step 3: Calculate the shortest distance
The formula for the shortest distance \( S.D. \) between two parallel lines is:\[\text{S.D.} = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\]Compute the cross product \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \):\[(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
4 & 0 & -5
2 & 1 & 1 \end{vmatrix} = 5\hat{i} - 14\hat{j} + 4\hat{k}\]Compute the magnitude of the direction vector \( \vec{b} \):\[| \vec{b} | = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}\]Compute the magnitude of the cross product:\[|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{5^2 + (-14)^2 + 4^2} = \sqrt{25 + 196 + 16} = \sqrt{237}\]The shortest distance is:\[\text{S.D.} = \frac{\sqrt{237}}{\sqrt{6}} = \sqrt{\frac{237}{6}} = \sqrt{\frac{79}{2}} \, \text{units.}\]