Question

If from a point $P(a,b,c)$ perpendiculars $PA$ 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. $- bcx + cay + abz = 0$

Answer 1

Answer: (B)

$bcx + cay - abz = 0$

Answer 2

$A ( 0 , b , c )$ in yz-plane and $B ( a , 0 , c )$ in zx-plane. Plane through O is It passes through A and B.

$\therefore 0 p + q b + r c = 0$ and $p a + 0 q + r c = 0$

$\Rightarrow \frac { p } { b c } = \frac { q } { c a } = \frac { r } { - a b } = k$

$\Rightarrow p = b c k , q = c a k$ and $r = - a b k$

Hence required plane is $b c x + c a y - a b z = 0$ .