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Q.
If the plane $ax-by+cz=0$ contains the line $\frac{x - a}{a}=\frac{y - 2 d}{b}=\frac{z - c}{c},\left(b \neq 0\right)$ then $\frac{b}{d}$ is equal to
NTA AbhyasNTA Abhyas 2022
Solution:
Since plane $ax-by+cz=0$ contains the line $\frac{x - a}{a}=\frac{y - 2 d}{b}=\frac{z - c}{c}$
$So,point\left(\right.a,2d,c\left.\right)$ 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 $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right),$ we get
$b^{2}=2bd$
$b=2d$
$\frac{b}{d}=2$