Question

The equation of the plane passing through the straight line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}$ and perpendicular to plane $x+2y+z=12$ is:

• A. $9x+2y-5z+8=0$
• B. $9x+2y-5z+10=0$
• C. $9x-2y+5z+6=0$
• D. $9x-2y-5z+4=0$

Answer 1

Answer: (D)

$9x-2y-5z+4=0$

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

Since it passes through a line.

$\therefore$ Normal of the plane must be perpendicular to the line

$\therefore a+2b+c=0$ ...... $\left(1\right)$ ($\perp$ to another plane)

$2a-b+4c=0$ ...... $\left(2\right)$

On solving :

$\frac{a}{8+1}=\frac{-b}{4-2}=\frac{c}{-1-4}$

$\Rightarrow \frac{a}{9}=\frac{b}{-2}=\frac{c}{-5}$

$a=9,b=-2,c=-5$

Any point on the straight line

$(2\alpha +1,-\alpha -1,4\alpha +3)$

putting $\alpha =1$

$(3,-2,7)$

$\therefore$ Equation of the line plane with normal DR $(9,-2,-5)$ and passing through $(3,-2,7)$

$\therefore 9(x-3)-2(y+2)-5(z-7)=0$

$\Rightarrow 9x-27-2y-4-5z+35=0$

$\Rightarrow 9x-2y-5z+4=0$