Question

If from a point P(a,b,c) perpendicular PA and PB are 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. None of these

Answer 1

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

Given that, PA and PB are perpendiculars from $P(a,b,c)$ to $yz$ and $zx$ planes.

On yz plane, $x=0$

Hence, $A=(0,b,c)$

Similarly, on zx plane, $y=0$,

Hence $B=(a,0,c)$

We need the equation of the plane passing through $(0,0,0)$ , $A$ & $B$ .

Equation of plane passing through $(0,0,0)$ is of form

$p(x-0)+q(y-0)+s(z-0)=0\Rightarrow px+qy+sz=0....eq.1$

As eq.1 passes through $(0,b,c)$ and $(a,0,c)$

$p\left(0\right)+q\left(b\right)+s\left(c\right)=0\Rightarrow qb+sc=0.....eq.2$

$p\left(a\right)+q\left(0\right)+s\left(c\right)=0\Rightarrow pa+sc=0.....eq.3$

Solving eq.2 and eq.3,

$\frac{p}{bc}=\frac{q}{ac}=\frac{s}{-ab}=k\Rightarrow p=kbc,q=kac,s=-kab$

Putting values of p, q, s in eq.1, we get

$kbc\left(x\right)+kac\left(y\right)-kab\left(z\right)=0\Rightarrow k(bcx+acy-abz)=0$

$\therefore bcx+cay-abz=0$ is the required equation.