The vector equation of line passing through (2, −1, 3) and perpendicular to the lines \(\displaystyle\frac{x-2}{3}=\frac{y-1}{1}=\frac{z+2}{2}\) and \(\displaystyle\frac{x+3}{-4}=\frac{y-5}{-3}=\frac{z+1}{2}\) is
(Here \(\lambda\) is a parameter)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\vec r= (2\hat{\imath}-\hat{\jmath}+3\hat{k}) +\lambda (8\hat{\imath}-14\hat{\jmath}-5\hat{k})\)
The denominators give the directions of the two given lines:
\(\vec b_1 = 3\hat{\imath} + \hat{\jmath} + 2\hat{k}\)
\(\vec b_2 = -4\hat{\imath} - 3\hat{\jmath} + 2\hat{k}\)
A line perpendicular to both is parallel to the cross product \(\vec b_1 \times \vec b_2\):
\( \displaystyle \vec b_1 \times \vec b_2 = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\3&1&2\\-4&-3&2\end{vmatrix} \)
\(= (2 + 6)\hat{\imath} - (6 + 8)\hat{\jmath} + (-9 + 4)\hat{k}\)
\(= 8\hat{\imath} - 14\hat{\jmath} - 5\hat{k}\)
The line through the point (2, −1, 3), with position vector \(2\hat{\imath} - \hat{\jmath} + 3\hat{k}\), in this direction is
\(\vec r = (2\hat{\imath} - \hat{\jmath} + 3\hat{k}) + \lambda(8\hat{\imath} - 14\hat{\jmath} - 5\hat{k})\)
Why the other options are incorrect
Option (2) \(\vec r= (-2\hat{\imath}+\hat{\jmath}-3\hat{k}) +\lambda (8\hat{\imath}-14\hat{\jmath}-5\hat{k})\) — Incorrect. The direction is right, but the line passes through (−2, 1, −3), not (2, −1, 3), and these two lines are different.
Option (3) \(\vec r= (-8\hat{\imath}-14\hat{\jmath}-5\hat{k}) +\lambda (-2\hat{\imath}+\hat{\jmath}+3\hat{k})\) — Incorrect. The direction \(-2\hat{\imath}+\hat{\jmath}+3\hat{k}\) is not perpendicular to \(\vec b_1\): the dot product is \(-6 + 1 + 6 = 1\).
Option (4) \(\vec r= (8\hat{\imath}-14\hat{\jmath}-5\hat{k}) +\lambda (2\hat{\imath}-\hat{\jmath}+3\hat{k})\) — Incorrect. The point and the direction are interchanged; \(2\hat{\imath}-\hat{\jmath}+3\hat{k}\) is not perpendicular to \(\vec b_1\) (dot product \(6 - 1 + 6 = 11\)).