The equation of the line passing through the point (2, −1, 3) and perpendicular to both the lines \(\displaystyle\frac{x-1}{1}=\frac{y+2}{2}=\frac{3-z}{1}\) and \(\displaystyle\frac{x+1}{3}=\frac{1-y}{1}=\frac{z-2}{2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{x-2}{3}=\frac{y+1}{-5}=\frac{z-3}{-7}\)
In standard form, \(\frac{3-z}{1} = \frac{z-3}{-1}\) and \(\frac{1-y}{1} = \frac{y-1}{-1}\). So the direction vectors of the two lines are
\(\vec{b}_1 = \hat{\imath}+2\hat{\jmath}-\hat{k}\)
\(\vec{b}_2 = 3\hat{\imath}-\hat{\jmath}+2\hat{k}\)
A line perpendicular to both is parallel to the cross product \(\vec{b}_1 \times \vec{b}_2\):
\( \displaystyle \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \)
\( = \hat{\imath}(4-1)-\hat{\jmath}(2+3)+\hat{k}(-1-6)\)
\( = 3\hat{\imath}-5\hat{\jmath}-7\hat{k}\)
Check: \(3-10+7 = 0\) and \(9+5-14 = 0\). Through (2, −1, 3):
\( \displaystyle \frac{x-2}{3}=\frac{y+1}{-5}=\frac{z-3}{-7} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x-2}{3}=\frac{y+1}{1}=\frac{z-3}{-5}\) — Incorrect. This uses 1, 2, 1 and 3, 1, 2, read without making the coefficients of z and y equal to +1; with the true direction 1, 2, −1 the dot sum is \(3+2+5 = 10 \ne 0\).
Option (2) \(\displaystyle\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-3}{-7}\) — Incorrect. The middle term of the cross product has the wrong sign; with 1, 2, −1 the dot sum is \(3+10+7 = 20 \ne 0\).
Option (4) \(\displaystyle\frac{x+2}{3}=\frac{y-1}{-5}=\frac{z+3}{-7}\) — Incorrect. The direction is right, but this line passes through (−2, 1, −3), not (2, −1, 3).