Question

Find the equation of the plane passing through the point $(1,2,1)$ and perpendicular to the line joining the points $(1,4,2)$ and $(2,3,5)$. Also find the coordinates of the foot of perpendicular and the perpendicular distance of the point $(4,0,3)$ from the above found plane.

Answer 1

DR of the line passing through $(1,4,2)$ and $(2,3,5)=(1,-1,3)$

Since the plane is perpendicular to this line,

$\therefore$ DR of the normal of the plane must be parallel to the line

$\therefore$ DR of the normal of the plane $=(1,-1,3)$

Plane passing through $(1,2,1)$

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

$\Rightarrow x-1-y+2+3z-3=0$

Equation of the plane; $\Rightarrow x-y+3z-2=0$

Perpendicular distance of $(4,0,3)$ from the $x-y+3z-2=0$

$\frac{|4-0+3(3)-2|}{\sqrt{1^2+1^2+3^2}}$

$=\frac{11}{\sqrt{11}}$

$=\sqrt{11}$ units

Equation of the line passing through $(1,4,2)$ and$(2,3,5)$

$=\frac{x-1}{1}=\frac{y-4}{-1}=\frac{z-2}{3}$

Any point on line $(\alpha +1,-\alpha +4,3\alpha +2)$

DR of the line with $(1,2,1)$

$=(\alpha ,\alpha -2,3\alpha -1)$

$\therefore$ DR will have same ratio

$\frac{\alpha }{1}=\frac{\alpha -2}{-1}=\frac{3\alpha -1}{3}$

$\alpha =1$

points = $(2,3,5)$