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

The correct option is D there is no such $\lambda$

The point $(\lambda ,1,3)$ which lies on the line, must lies on the plane.
$\Rightarrow \lambda -2=0\Rightarrow \lambda =2$

We also know that when a line lies on the plane, then the direction ratios of the line must be perpendicular to the direction ratios of the normal to the plane.
$\therefore$The line must be perpendicular to the normal to the plane.
$\Rightarrow 3\times 1+(2+\lambda )(-2)+(-1)\times 0=0\Rightarrow \lambda =-\frac{1}{2}$
In both the cases, we get different values of $\lambda$
Thus, there is no value of $\lambda$ for which the line can lie on the given plane.