The shortest distance between the lines \(\vec{r} = (\hat{\jmath}-\hat{k})+\lambda(\hat{\imath}+2\hat{\jmath}+2\hat{k})\) and \(\vec{r} = (p\hat{\imath}+4\hat{\jmath})+\mu(2\hat{\imath}+\hat{\jmath}-2\hat{k})\) is 1. Then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(p = 1\) or \(p = 4\)
Here
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 1 & -2 \end{vmatrix}\)
\(= \hat{\imath}(-4-2)-\hat{\jmath}(-2-4)+\hat{k}(1-4) = -6\hat{\imath}+6\hat{\jmath}-3\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{36+36+9} = 9\)
\(\vec{a}_2-\vec{a}_1 = p\hat{\imath}+3\hat{\jmath}+\hat{k}\)
\((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = -6p+18-3 = 15-6p\)
The shortest distance is 1:
\( \displaystyle \frac{|15-6p|}{9} = 1 \)
\(|15-6p| = 9\)
Because of the absolute value, there are two cases:
\(15-6p = 9\) gives \(p = 1\)
\(15-6p = -9\) gives \(p = 4\)
Why the other options are incorrect
Option (1) \(p = -1\) or \(p = -4\) — Incorrect. For \(p = -1\), \(\frac{|15+6|}{9} = \frac{7}{3} \ne 1\).
Option (2) \(\displaystyle p = \frac{5}{2}\) only — Incorrect. For \(p = \frac{5}{2}\), \(15-6p = 0\): the lines then intersect and the shortest distance is 0.
Option (4) \(p = 1\) or \(p = -4\) — Incorrect. For \(p = -4\), \(\frac{|15+24|}{9} = \frac{13}{3} \ne 1\).