Question

The length of the perpendicular from the origin to the plane $2x+3y+\lambda z=1$ $(\lambda >0)$ is $\frac{1}{5}$. Then $\lambda$ is

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $0$
• D. $1$

Answer 1

The correct option is A $2\sqrt{3}$

The equation of line of perpendicular from the origin to the plane $P:2x+3y+\lambda z=1$ is $L:\frac{x}{2}=\frac{y}{3}=\frac{z}{\lambda }$
Let the foot of perpendicular of line to plane be $A(2t,3t,\lambda t)$
Since, it lies on plane
Therefore, $2\left(2t\right)+3\left(3t\right)+\lambda \left(\lambda t\right)=1$
$\Rightarrow t=\frac{1}{13+{\lambda }^2}$
Given that $OA=\frac{1}{5}$
Therefore, $\sqrt{4t^2+9t^2+{\lambda }^2{t}^2}=\frac{1}{5}$
$\Rightarrow \left(\frac{1}{13+{\lambda }^2}\right)\sqrt{13+{\lambda }^2}=\frac{1}{5}$
$\Rightarrow 13+{\lambda }^2=25$
$\Rightarrow \lambda =2\sqrt{3}$
Ans: A