The vector equation of line passing through \((-1,3,-2)\) and perpendicular to the lines \(\displaystyle\frac{x+4}{1}=\frac{y}{2}=\frac{z-3}{3}\) and \(\displaystyle\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-6}{5}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\vec{r}=(-\hat{\imath}+3\hat{\jmath}-2\hat{k})+\lambda(2\hat{\imath}-7\hat{\jmath}+4\hat{k})\)
A line perpendicular to both given lines is parallel to the cross product of their direction vectors:
\(\vec{b}_1 = \hat{\imath} + 2\hat{\jmath} + 3\hat{k}\)
\(\vec{b}_2 = -3\hat{\imath} + 2\hat{\jmath} + 5\hat{k}\)
\( \displaystyle \vec{b}_1\times\vec{b}_2 = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 2 & 3\\ -3 & 2 & 5\end{vmatrix} \)
\(= (10 - 6)\hat{\imath} - (5 + 9)\hat{\jmath} + (2 + 6)\hat{k}\)
\(= 4\hat{\imath} - 14\hat{\jmath} + 8\hat{k}\)
\(= 2(2\hat{\imath} - 7\hat{\jmath} + 4\hat{k})\)
So the direction is \(2\hat{\imath} - 7\hat{\jmath} + 4\hat{k}\). The line passes through \((-1, 3, -2)\), with position vector \(-\hat{\imath} + 3\hat{\jmath} - 2\hat{k}\):
\(\vec{r} = (-\hat{\imath} + 3\hat{\jmath} - 2\hat{k}) + \lambda(2\hat{\imath} - 7\hat{\jmath} + 4\hat{k})\)
Shortcut: The direction in the right option has a zero dot product with both \(\vec{b}_1\) and \(\vec{b}_2\):
Options (1), (3): \((-1)(1) + 3(2) + (-2)(3) = -1 \ne 0\)
Option (2): \((-3)(1) + 4(2) + 15(3) = 50 \ne 0\)
Option (4): \(2(1) - 7(2) + 4(3) = 0\) and \(2(-3) - 7(2) + 4(5) = 0\)
Only Option (4) has a direction perpendicular to both lines, and it passes through \((-1, 3, -2)\).