The direction cosines of the line which is perpendicular to the lines…

The correct answer is Option (1) → \(\displaystyle\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\)

A line perpendicular to two lines is along the cross product of their direction vectors:

    \( \displaystyle \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & -2 & -2\\ 0 & 2 & 1\end{vmatrix} \)

    \( = \hat{\imath}(-2+4) - \hat{\jmath}(1-0) + \hat{k}(2-0) \)

    \( = 2\hat{\imath}-\hat{\jmath}+2\hat{k} \)

So the direction ratios are 2, −1, 2. Dividing by \(\sqrt{2^2+1^2+2^2} = 3\) gives the direction cosines:

    \( \displaystyle \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \)

Shortcut: The required direction must have zero dot product with both given sets. Using 3 × each option:

    Option (1): \((2)(1)+(-1)(-2)+(2)(-2) = 0\) and \((-1)(2)+(2)(1) = 0\)
    Option (2): \((-2)(1)+(-1)(-2)+(2)(-2) = -4 \ne 0\)
    Option (3): \((2)(1)+(-1)(-2)+(-2)(-2) = 8 \ne 0\)
    Option (4): \((2)(1)+(1)(-2)+(2)(-2) = -4 \ne 0\)

Only Option (1) is perpendicular to both.