Question

The foot of the perpendicular from the point of intersection of line $\frac{x-3}{-1}=\frac{y-2}{1}=\frac{z}{5}$ and the plane $x-y+2z=9$ on the plane $3x+4y-z=0$ is:

• A. $(\frac{1}{2},1,\frac{11}{2})$
• B. $(\frac{3}{2},1,\frac{-17}{2})$
• C. $(\frac{5}{2},1,\frac{-9}{2})$
• D. $(\frac{3}{2},1,\frac{-11}{2})$

Answer 1

The correct option is A $(\frac{1}{2},1,\frac{11}{2})$

Any point on the given line $\frac{x-3}{-1}=\frac{y-2}{1}=\frac{z}{5}$ can be taken as $(x,y,z)=(3-t,2+t,5t)$
Now for intersection with the given plane, $(3-t,2+t,5t)$ must lie on the plane $x-y+2z=9$
Hence, $(3-t)-(2+t)+2\left(5t\right)=9$
$\Rightarrow 8t=8\Rightarrow t=1$
Hence the point of intersection is $(3-1,2+1,5\times 1)$ i.e. $(2,3,5)$
Foot of perpandicular from $(x_1,y_1,z_1)=(2,3,5)$ to plane $3x+4y-z=0$ is given by :
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{a{x}_1+b{y}_1+c{z}_1-d}{a^2+b^2+c^2}\Rightarrow \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{-1}=-\frac{6+12-5}{9+16+1}\Rightarrow (x,y,z)=(\frac{1}{2},1,\frac{11}{2})$