The shortest distance between the lines \(\vec{r} = (1+2t)\hat{\imath}+(2t-3)\hat{\jmath}+(4-t)\hat{k}\) and \(\vec{r} = (s+3)\hat{\imath}-(2s+2)\hat{\jmath}+(7-2s)\hat{k}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
Separate the terms without the parameter from those with it, writing each line as \(\vec{a}+t\vec{b}\):
\(\vec{r} = (\hat{\imath}-3\hat{\jmath}+4\hat{k})+t(2\hat{\imath}+2\hat{\jmath}-\hat{k})\)
\(-(2s+2) = -2-2s\), so
\(\vec{r} = (3\hat{\imath}-2\hat{\jmath}+7\hat{k})+s(\hat{\imath}-2\hat{\jmath}-2\hat{k})\)
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 2 & -1 \\ 1 & -2 & -2 \end{vmatrix}\)
\(= \hat{\imath}(-4-2)-\hat{\jmath}(-4+1)+\hat{k}(-4-2) = -6\hat{\imath}+3\hat{\jmath}-6\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{36+9+36} = 9\)
\(\vec{a}_2-\vec{a}_1 = 2\hat{\imath}+\hat{\jmath}+3\hat{k}\)
\((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = -12+3-18 = -27\)
By the shortest distance formula:
\( \displaystyle d = \frac{|-27|}{9} = 3 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{5}{\sqrt{17}}\) — Incorrect. This comes from taking the second direction as \(\hat{\imath}+2\hat{\jmath}-2\hat{k}\), losing the minus sign in front of \((2s+2)\).
Option (3) \(\displaystyle\frac{5}{3}\) — Incorrect. This comes from taking the second point as (3, 2, 7), again losing the minus sign in front of \((2s+2)\); then the dot product is \(-15\).
Option (4) 27 — Incorrect. This is only the numerator, not divided by \(|\vec{b}_1\times\vec{b}_2| = 9\).