Question

The value of $\lambda$ for which the straight line $\frac{x-\lambda }{3}=\frac{y-1}{2+\lambda }=\frac{z-3}{-1}$ may lie on the plane $x-2y=0$ is

• A. $2$
• B. $0$
• C. $-\frac{1}{2}$
• D. There is no such $\lambda$

Answer 1

Answer: (C)

$-\frac{1}{2}$

For the line $\frac{x-\lambda }{3}=\frac{y-1}{2+\lambda }=\frac{z-3}{-1}$ to lie on the plane, then this line must be perpendicular to the normal of plane $x-2y=0$

Line vector, $L:3\hat{i}+(2+\lambda )\hat{j}-\hat{k}$

Normal of plane, $N:\hat{i}-2\hat{j}$

To satisfy the condition of perpendicularity,

$L\cdot N=0$

$\Rightarrow (3\hat{i}+(2+\lambda )\hat{j}-\hat{k})\cdot (\hat{i}-2\hat{j})=0$

$\Rightarrow 3\cdot 1-(2+\lambda )\cdot 2=0$

$\Rightarrow 3-4-2\lambda =0$

$\Rightarrow \lambda =-\frac{1}{2}$

Hence, option C.