Question

If the plane $ax-by+cz=0$ contains the line $\frac{x-a}{a}=\frac{y-2d}{b}=\frac{z-c}{c},\left(b\ne 0\right)$ then $\frac{b}{d}$ is equal to

• A. $1$
• B. $2$
• C. $0$
• D. $-3$

Answer 1

Answer: (B)

Since plane $ax-by+cz=0$ contains the line $\frac{x-a}{a}=\frac{y-2d}{b}=\frac{z-c}{c}$
$So,point(a,2d,c)$ will satisfy the plane and normal of plane will make $90^{\circ }$ with line
So, $a\cdot a-b\cdot 2d+c^2=0$
$c^2+a^2=2bd...\left(1\right)$
and $a^2-b^2+c^2=0$ , since angle between normal of plane and line is $90^{\circ }$
$b^2=a^2+c^2...\left(2\right)$
From equation $(1)$ and $(2),$ we get
$b^2=2bd$
$b=2d$
$\frac{b}{d}=2$