The two lines given by \(\vec{r}=\hat{\imath}+\hat{\jmath}+\lambda(2\hat{\imath}-\hat{\jmath}+\hat{k})\) and \(\vec{r}=(2\hat{\imath}+\hat{\jmath}-\hat{k})+\mu(4\hat{\imath}-2\hat{\jmath}+2\hat{k})\), which of the following are true about these two lines.
(A) These lines are perpendicular
(B) These lines are parallel
(C) The shortest distance between given lines is zero
(D) The shortest distance between given lines is \(\displaystyle\frac{\sqrt{11}}{\sqrt{6}}\)
(E) The shortest distance between given lines is \(\sqrt{11}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
The direction vectors are
\(\vec{b}_1 = 2\hat{\imath}-\hat{\jmath}+\hat{k}\)
\(\vec{b}_2 = 4\hat{\imath}-2\hat{\jmath}+2\hat{k} = 2\vec{b}_1\)
So the lines are parallel. For parallel lines with \(\vec{b} = \vec{b}_1\), \(\vec{a}_1 = \hat{\imath}+\hat{\jmath}\) and \(\vec{a}_2 = 2\hat{\imath}+\hat{\jmath}-\hat{k}\), the shortest distance is
\( \displaystyle d = \frac{\left|\vec{b}\times(\vec{a}_2-\vec{a}_1)\right|}{|\vec{b}|} \)
\(\vec{a}_2-\vec{a}_1 = \hat{\imath}-\hat{k}\)
\( \displaystyle \vec{b}\times(\vec{a}_2-\vec{a}_1) = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & -1 & 1\\ 1 & 0 & -1\end{vmatrix} \)
\(= (1-0)\hat{\imath}-(-2-1)\hat{\jmath}+(0+1)\hat{k}\)
\(= \hat{\imath}+3\hat{\jmath}+\hat{k}\)
\( \displaystyle d = \frac{\sqrt{1+9+1}}{\sqrt{4+1+1}} = \frac{\sqrt{11}}{\sqrt{6}} \)
So the lines are parallel (not perpendicular, since \(\vec{b}_1\cdot\vec{b}_2 = 8+2+2 = 12 \ne 0\)), and the shortest distance between them is \(\frac{\sqrt{11}}{\sqrt{6}}\).
Checking the statements
(A) These lines are perpendicular — Incorrect. \(\vec{b}_1\cdot\vec{b}_2 = 12 \ne 0\).
(B) These lines are parallel — Correct. \(\vec{b}_2 = 2\vec{b}_1\).
(C) The shortest distance between given lines is zero — Incorrect. It is \(\frac{\sqrt{11}}{\sqrt{6}}\); the lines are distinct parallel lines.
(D) The shortest distance between given lines is \(\frac{\sqrt{11}}{\sqrt{6}}\) — Correct. \(\left|\vec{b}\times(\vec{a}_2-\vec{a}_1)\right| = \sqrt{11}\) divided by \(|\vec{b}| = \sqrt{6}\).
(E) The shortest distance between given lines is \(\sqrt{11}\) — Incorrect. \(\sqrt{11}\) still has to be divided by \(|\vec{b}| = \sqrt{6}\).