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 $\lambda$

If a line $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ lies on a plane $a_{2}x+b_{2}y+c_{z}z=d$, then
i) $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
ii) $a_2{x}_1+b_2{y}_1+c_2{z}_1=d$
According to the question,
$3\left(1\right)+(2+\lambda )(-2)+(-1)\left(0\right)=0$
$\Rightarrow 3-4-2\lambda =0$
$\Rightarrow \lambda =-1/2$
Also, $\lambda -2=0$
$\Rightarrow \lambda =2$
So, there is no such value of $\lambda$.