The shortest distance between the lines \(\displaystyle\frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{-2}\) and \(\displaystyle\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{11}{3}\)
Reading each line in the form \(\displaystyle\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\):
\(\vec{a}_1 = \hat{\imath}-2\hat{\jmath}+3\hat{k},\quad \vec{b}_1 = 2\hat{\imath}+\hat{\jmath}-2\hat{k}\)
\(\vec{a}_2 = -\hat{\imath}+\hat{\jmath}+2\hat{k},\quad \vec{b}_2 = \hat{\imath}+2\hat{\jmath}+2\hat{k}\)
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{vmatrix}\)
\(= \hat{\imath}(2+4)-\hat{\jmath}(4+2)+\hat{k}(4-1) = 6\hat{\imath}-6\hat{\jmath}+3\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{36+36+9} = 9\)
\(\vec{a}_2-\vec{a}_1 = -2\hat{\imath}+3\hat{\jmath}-\hat{k}\)
\((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = -12-18-3 = -33\)
By the shortest distance formula:
\( \displaystyle d = \frac{|-33|}{9} = \frac{11}{3} \)
Why the other options are incorrect
Option (1) 33 — Incorrect. This is only the numerator, not divided by \(|\vec{b}_1\times\vec{b}_2| = 9\).
Option (2) \(\sqrt{14}\) — Incorrect. This is \(|\vec{a}_2-\vec{a}_1|\), the distance between the two given points, which is not the shortest distance.
Option (3) 11 — Incorrect. This divides 33 by \(|\vec{b}_1| = 3\) instead of \(|\vec{b}_1\times\vec{b}_2| = 9\).