Question

If $A$ and $B$ are foot of perpendicular drawn from point $Q (a, b, c)$ to the planes $yz$ and $zx$, then equation of plane through the points $A, B$ and $O$ is ___________

• A. $\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$
• B. $\frac{x}{a}-\frac{y}{b} + \frac{z}{c}=0$
• C. $\frac{x}{a} - \frac{y}{b} - \frac{z}{c}=0$
• D. $\frac{x}{a} + \frac{y}{b} + \frac{z}{c}=0$

Answer 1

Answer: (A)

$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$

Answer 2

The foot of perpendicular from point $Q(a, b, c)$ to the $yz$ plane is $A(0, b, c)$ and the foot of perpendicular from point $Q$ to the $zx$ plane in $B(a, 0, c)$
Let the equation of plane passing through the point $(0,0,0)$ be
$Ax+By+Cz=0\, \dots(i)$
Also it is paring through the point $A(0, b, c)$ and $B(a, 0, c)$
$\therefore 0+Bb+Cc=0$
and $Aa+0+Cc =0$
$\Rightarrow Cc=Bb$
and $Cc=-Aa$
$\therefore -A a =-B b=C c=k$
$\Rightarrow A=-\frac{k}{a}, B=-\frac{k}{b}$
and $C =\frac{k}{c}$
From E (i), $-\frac{k}{a} x-\frac{k}{b} y+\frac{k}{c} z=0$
$\Rightarrow -\frac{x}{a}-\frac{y}{b}+\frac{z}{c}=0$ or $\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$