Question

The equation of plane passing through a point $A(2,-1,3)$ and parallel to the vectors $a=(3,0,-1)$ and $b=(-3,2,2)$ is:

• A. $2x-3y+6z-25=0$
• B. $2x-3y+6z+25=0$
• C. $3x-2y+6z-25=0$
• D. $3x-2y+6z+25=0$

Answer 1

Answer: (A)

$2x-3y+6z-25=0$

Let the DR of the normal of required plane be $<a,b,c>$

Since plane is parallel to $(3,0,-1)$

$\therefore$ normal must be perpendicular

$\therefore 3a+0b-c=0$ ... $\left(1\right)$

Also, it is parallel to $(-3,2,2)$

$\therefore -3a+2b+2c=0$ ... $\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\frac{a}{2}=\frac{-b}{6-3}=\frac{c}{6}$

$a=2,b=-3,c=6$

It passes through $(2,-1,3)$

$2(x-2)-3(y+1)+6(z-3)=0$

$\Rightarrow 2x-4-3y-3+6z-18=0$

$\Rightarrow 2x-3y+6z-25=0$