Question

The projection of the line $\frac{x-2}{4}=\frac{y+2}{8}=\frac{z-3}{6}$ in the plane $P_1:2x+3y-4z+6=0$ is the line on the intersection of planes $P_1$ and $P_2$. If $P_1$ is perpendicular to $P_2$, then equation of the plane $P_2$ is

• A. $25x-14y+2z=84$
• B. $25x-14y+2z=-34$
• C. $-25x-14y+2z=84$
• D. $25x+14y+2z=34$

Answer 1

The correct option is A $25x-14y+2z=84$

$\because P_1\perp P_2$
$\Rightarrow P_2$ contains the given line and its projection also.
Let direction ratios of the normal of the plane $P_2$ be $a,b,c$.
$P_2$ contains the line $\frac{x-2}{4}=\frac{y+2}{8}=\frac{z-3}{6}$, so it passes through $(2,-2,3)$.
Let equation of $P_2$ be $a(x-2)+b(y+2)+c(z-3)=0$.

Since $P_2$ is perpendicular to $P_1$ as well as the given line,
$\therefore 4a+8b+6c=0\cdots \left(1\right)$
and $2a+3b-4c=0\cdots \left(2\right)$
$\left(1\right)-2\times \left(2\right)$, we get
$b=-7c$
$\therefore a=\frac{25c}{2}$
Hence, equation of the plane $P_2$ is
$\frac{25c}{2}(x-2)-7c(y+2)+c(z-3)=0$
i.e., $25x-14y+2z=84$