Question

A plane which contain the line $\frac{y}{b}+\frac{z}{c}=1,x=0$, and parallel to the line $\frac{x}{a}-\frac{z}{c}=1,y=0$. If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the direction cosines of the a line $l$ then distance of the plane from the origin is

Answer 1

The equation of the line contained in the required plane
$\frac{y}{b}+\frac{z}{c}=1,x=0\frac{y}{b}-\frac{1}{2}=\frac{1}{2}-\frac{z}{c},x=0\Rightarrow \frac{y-\frac{1}{2}b}{b}=\frac{z-\frac{1}{2}c}{-c}=\frac{x}{0}...\left(1\right)$
Equation of the line which is parallel to plane
$\frac{x-\frac{1}{2}a}{a}=\frac{z+\frac{1}{2}c}{c}=\frac{y}{0}...\left(2\right)$
Equation of the plane containing the line $\left(1\right)$ is
$A(x-0)+B\left(y-\frac{1}{2}b\right)+C\left(z-\frac{1}{2}c\right)=0\text{here}A.0+B.b-C.c=\mathrm{0...}\left(3\right)$
If the plane parallel to the plane, then
$A.a+B.0+C.c=\mathrm{0...}\left(4\right)$
From equation $\left(3\right)$ and $\left(4\right)$
$\frac{A}{bc}=\frac{B}{-ac}=\frac{C}{-ab}$
putting the value of the $A,B,C$ in the equation of the plane
$bcx-ac\left(y-\frac{1}{2}b\right)-ab\left(z-\frac{1}{2}c\right)=0\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=1$
Distance of the plane fron the origin is
$\frac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=1\because \frac{1}{a},\frac{1}{b},\frac{1}{c}$
are the direction cosines of a line
$\therefore \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1$