Question

If a point $P(a,b,c)$ perpendiculars $PA,PB$ and drawn to $YZ$ and $ZX$ planes, then the equation of the plane $OAB$ is

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

Answer 1

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

The coordinates of $A$ and $B$ are $(0,b,c)$ and $(a,0,c)$ respectively.
The equation of a plane passing through $O(0,0,0)$ is
$Px+Qy+Rz=\mathrm{0....}\left(i\right)$
It passes through $A$ and $B$
$\therefore P\times 0+Q\times b+R=0$
and $P\times a+Q\times 0+R\times c=0$
$\Rightarrow \frac{P}{bc}=\frac{Q}{ac}=\frac{R}{-ab}$
Substituting the values of $P,Q,R$ in Eq. (i) we get
$bcx+acy-abz=0$