The equation of line passing through the point (2, 1, 3) and perpendicular to the lines \( \displaystyle\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \) and \( \displaystyle\frac{x}{-3}=\frac{y}{2}=\frac{z}{5} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{x-2}{2}=\frac{1-y}{7}=\frac{z-3}{4} \)
A line perpendicular to both lines is parallel to the cross product of their direction vectors \( \vec{b}_1=\hat{\imath}+2\hat{\jmath}+3\hat{k} \) and \( \vec{b}_2=-3\hat{\imath}+2\hat{\jmath}+5\hat{k} \).
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & 2 & 3\\-3 & 2 & 5\end{vmatrix} \)
\( \displaystyle =(10-6)\hat{\imath}-(5+9)\hat{\jmath}+(2+6)\hat{k} \)
\( \displaystyle =4\hat{\imath}-14\hat{\jmath}+8\hat{k} \)
So the direction ratios can be taken as 2, −7, 4. The line through (2, 1, 3) is
\( \displaystyle \frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4} \)
Since \( \frac{y-1}{-7}=\frac{1-y}{7} \), this is the same as
\( \displaystyle \frac{x-2}{2}=\frac{1-y}{7}=\frac{z-3}{4} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{x+2}{2}=\frac{y+1}{-7}=\frac{z+3}{4} \) — Incorrect. This line passes through (−2, −1, −3), not through (2, 1, 3).
Option (3) \( \displaystyle\frac{x-2}{+2}=\frac{y-1}{-7}=\frac{z-3}{-4} \) — Incorrect. Its direction ratios are 2, −7, −4; the k-component of the cross product is 1(2) − 2(−3) = 8, so the last ratio must be +4.
Option (4) \( \displaystyle\frac{x+2}{2}=\frac{y+1}{-7}=\frac{z+3}{-4} \) — Incorrect. This line passes through (−2, −1, −3), and its last direction ratio also has the wrong sign.