Question

lf the planes $x+2y-z+5=0,2x-ky+4z+3=0$ are perpendicular, then $k$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is B $-1$

I am using the constant $\lambda$ instead of $k$ to avoid the confusion.

The normals to the planes are given by $i+2j-k$ and $2i-\lambda j+4k$, respectively.

Since, they are perpendicular dot product between normals are zero.

Thus, $(i+2j-k).(2i-\lambda +4k)=0$.

$\Rightarrow 2-2\lambda -4=0\Rightarrow \lambda =-1$