Consider the lines \[ \begin{aligned} & L_{1}: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2} \\ & L_{2}: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} \end{aligned} \] The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

(B) $\frac{-\hat{i}-7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$

(A) $\frac{-\hat{i}+7 \hat{j}+7 \hat{k}}{\sqrt{99}}$

(C) $\frac{-\hat{i}+7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$

(D) $\frac{7 \hat{i}-7 \hat{j}-\hat{k}}{\sqrt{99}}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Vector Algebra

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2} L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The unit vector perpendicular to both

L_{1}

and

L_{2}

Answer: (B)

The direction vectors are

d_{1} = 3 \hat{i} + \hat{j} + 2 \hat{k}

and

d_{2} = \hat{i} + 2 \hat{j} + 3 \hat{k}

. The cross product is

d_{1} \times d_{2} = \hat{i} 31 \hat{j} 12 \hat{k} 23 = - \hat{i} - 7 \hat{j} + 5 \hat{k}

. The magnitude is

1 + 49 + 25 = 75 = 53

. So the unit vector is

\frac{- i ^ - 7 j ^ + 5 k ^}{5 3}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option