The shortest distance between the lines \(\displaystyle\frac{1-x}{2}=\frac{y-3}{1}=\frac{z+1}{2}\) and \(\displaystyle\frac{2x-4}{4}=\frac{y+1}{2}=\frac{z-3}{1}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{17}{3}\)
The direction ratios can be read only in the standard form, with x, y and z having coefficient 1:
\( \displaystyle \frac{1-x}{2} = \frac{x-1}{-2},\qquad \frac{2x-4}{4} = \frac{x-2}{2} \)
So
\(\vec{a}_1 = \hat{\imath}+3\hat{\jmath}-\hat{k},\quad \vec{b}_1 = -2\hat{\imath}+\hat{\jmath}+2\hat{k}\)
\(\vec{a}_2 = 2\hat{\imath}-\hat{\jmath}+3\hat{k},\quad \vec{b}_2 = 2\hat{\imath}+2\hat{\jmath}+\hat{k}\)
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}\)
\(= \hat{\imath}(1-4)-\hat{\jmath}(-2-4)+\hat{k}(-4-2) = -3\hat{\imath}+6\hat{\jmath}-6\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{9+36+36} = 9\)
\(\vec{a}_2-\vec{a}_1 = \hat{\imath}-4\hat{\jmath}+4\hat{k}\)
\((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = -3-24-24 = -51\)
By the shortest distance formula:
\( \displaystyle d = \frac{|-51|}{9} = \frac{17}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{\sqrt{17}}\) — Incorrect. This comes from reading the first direction as \(2\hat{\imath}+\hat{\jmath}+2\hat{k}\), missing that \(\frac{1-x}{2} = \frac{x-1}{-2}\).
Option (2) 17 — Incorrect. This divides 51 by \(|\vec{b}_1| = 3\) instead of \(|\vec{b}_1\times\vec{b}_2| = 9\).
Option (3) 51 — Incorrect. This is only the numerator, not divided by \(|\vec{b}_1\times\vec{b}_2| = 9\).