Question

The equation of the plane through $(4,4,0)$ and perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$ is

• A. 4x + 3y + 3z = 28
• B. 4x - 2y - 3z = 8
• C. 4x + 2y + 3z = 24
• D. 4x + 2y - 3z = 24

Answer 1

Answer: (B)

Equations of plane passing through $(4,4,0)$ is given by $a(x-4)+b(y-4)+c(z-0)=0$, where $a,b,c$ are DR's of normal to the plane.
Since this plane is $\perp$ to the given plans, therefore, we get
$2a+b+2c=0$
and $3a+3b+2c=0$
By cross-multiplication method
$\frac{a}{2-6}=\frac{-b}{4-6}=\frac{c}{6-3}$
$\Rightarrow \frac{a}{-4}=\frac{b}{2}=\frac{c}{3}$
So, the required equation of plane is
$-4(x-4)+2(y-4)+3(z)=0$
$\Rightarrow -4x+16+2y-8+3z=0$
$\Rightarrow 4x-2y-3z=8$