Question

If the plane $4(x-1)+k(y-2)+8(z-5)=0$ contains the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-5}{3}$, then $k$ is

• A. $2$
• B. $4$
• C. $-8$
• D. $8$

Answer 1

The correct option is A $2$

Given that,

plane $4(x-1)+k(y-2)+8(2-5)=0$ contain the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-5}{3}$.

We know that,

if plane Ax + Bu + Cz + D = 0 contains a line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c},$ then aA + bB + cC = 0 and $A{x}_1+B{y}_1+C{z}_1+D=0$

Here, plane is $4x-4+ky-2k+8z-40=0\Rightarrow 4x+ky+8z-44-2k=0$

$\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-5}{3}$

$\Rightarrow 4\left(2\right)+k\left(4\right)+8\left(3\right)-44-2k=0\Rightarrow 8+4k+24-44-2k=0\Rightarrow 2k=12\Rightarrow k=6$