Question

Find the direction ratios of the normal to the plane passing through the point (2,1,3) and the line of intersection of the planes x+2y+z=3 and 2x-y-z=5.

Answer 1

The equation of the plane passing through the intersection of the planes $x+2y+z=3$ and $2x-y-z=5$ is given by

$(x+2y+z-3)+\lambda (2x-y-z-5)=0$

$x(2\lambda +1)+y(2-\lambda )+z(1-\lambda )-3-5\lambda =0$

It passes through (2,1,3). Therefore,

$2(2\lambda +1)+(2-\lambda )+3(1-\lambda )-3-5\lambda =0$

$4\lambda +2+2-\lambda +3-3\lambda -3-5\lambda =0$

$4-5\lambda =0$

$\lambda =\frac{4}{5}$

Substituting this value in equation of plane, we get,

$13x+6y+z-35=0$ as the equation of required plane.

Direction ratios of the normal to this plane are proportional to 13,6,1.