A line is perpendicular to two lines having direction ratios 1, −2, −2 and 0, 2, 1 then the direction cosines of the line are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{2}{3},-\frac{1}{3},\frac{2}{3} \)
A line perpendicular to both lines is parallel to the cross product of their direction vectors:
\( \displaystyle \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -2 & -2\\0 & 2 & 1\end{vmatrix} \)
\( \displaystyle =(-2+4)\hat{\imath}-(1-0)\hat{\jmath}+(2-0)\hat{k} \)
\( \displaystyle =2\hat{\imath}-\hat{\jmath}+2\hat{k} \)
So the direction ratios are 2, −1, 2, and
\( \displaystyle \sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3 \)
Dividing by 3, the direction cosines are \( \pm\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right) \); Option (3) is the one with the + sign.
Shortcut: Take the dot product of each option with (1, −2, −2) and (0, 2, 1). Only Option (3) gives 0 with both: \( \frac{2}{3}+\frac{2}{3}-\frac{4}{3}=0 \) and \( 0-\frac{2}{3}+\frac{2}{3}=0 \).