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

Answer 1

Answer: (C)

Since, plane is parallel to a given vector, it implies that normal of plane must perpendicular to the given vectors. Given point to which plane passes through is $(2,-1,3)$. Let $A,B$ and $C$ are direction ratios of its normal.
$\therefore$ Equation of plane is
$A(x-2)+B(y+1)+C(z-3)=0$
Now, normal to plane $Ai+Bj+Ck$ is perpendicular to the given vectors
$a=3i+0j-k$
and $b=-3i+2j+2k$
$\therefore 3A+OB-C=0$ ...(i)
and $-3A+2B+2C=0$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{A}{2}=\frac{B}{-3}=\frac{C}{6}$
$\therefore$ Equation of plane be
$2(x-2)-3(y+1)+6(z-3)=0$
i.e.,$2x-3y+6z-25=0$