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

Answer 1

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

Given: $\frac{x-\lambda }{3}=\frac{y-1}{2+\lambda }=\frac{z-3}{-1}$​
So, the fixed point of the line $(\lambda ,1,3)$ must lie in the plane. Also the product of DR's of line and plane will be zero.
Condition $I$
Putting $(\lambda ,1,3)$ in the plane
We get $\lambda -2\times 1=0$
$\Rightarrow \lambda =2\cdots \left(i\right)$
Condition $II$
Product of DR's of line and plane
$\Rightarrow 1\left(3\right)-2(2+\lambda )+0(-1)=0$
$\Rightarrow 3-4-2\lambda$
$\Rightarrow \lambda =\frac{-1}{2}\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ we obtain different values of $\lambda$.
Hence, there is no such value of $\lambda$ for which the line lies on the plane.