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Q. Equation of the plane passing through $ (- 2, 2, 2) $ and $ (2, - 2 ,- 2 ) $ and perpendicular to the plane $ 9x - 13y - 3z = 0 $ is

MHT CETMHT CET 2009

Solution:

Any plane passing through $(-2,2,2)$ is
$A(x+2)+B(y-2)+C(z-2)=0$
$\because$ It passes through $(2,-2,-2)$
$\Rightarrow \,\,\,4 A-4 B-4 C=0 \,\,\,\,\,\,...(i)$
It is parallel to $9 x-13 y-3 z=0$
$\therefore \,\,\,9 A-13 B-3 C=0 \,\,\,\,\,.(ii) $
Solving Eqs. (i) and (ii),
$\frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36}$
$\Rightarrow \,\,\,\frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}$
$\therefore $ Required equation of plane is
$- 40(x+2)-24(y-2)-16(z-2)=0$
$\Rightarrow -40 x-80-24 y+48-16 z+32=0 $
$\Rightarrow \,\,\, 40 x+24 y+16 z=0$
$\Rightarrow \,\,\,5 x+3 y+2 z=0$