Question

From the point $P(a,b,c)$ the normals drawn to planes $yz$ and $zx$ are $PA,PB,$ then the equation of plane $OAB$ is

• A. $bcx+acy+abz=0$
• B. $bcx+acy-abz=0$
• C. $bcx-acy+abz=0$
• D. $-bcx+acy+abz=0$

Answer 1

The correct option is B $bcx+acy-abz=0$

$PA,PB$ are perpendicular drawn from $P(a,b,c)$ on $yz$ and $zx$ planes

$\therefore A(0,b,c)$ and $B(a,0,c)$ are points on $yz$ and $zx$ planes.

The equation of plane passing through $(0,0,0,)$ is

$Ax+By+Cz=0$

which also passes through $A$ and $B$

$\therefore A.0+B.b+C.c=0$ ....$\left(1\right)$

$A.a+B.0+C.c=0$ ....$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow \frac{A}{bc-0}=\frac{B}{ac-0}=\frac{C}{0-ab}=\lambda \Rightarrow A=\lambda bc,B=\lambda ac,C=-\lambda ab$

Required equation is $bcx+acy-abz=0$