Question

The locus of foot of the perpendiculars drawn from the line $\frac{x-2}{2}=\frac{y-1}{1}=\frac{z}{3}$ to the plane $x+y+z=1$ is

• A. $\frac{x-1}{0}=\frac{y}{-1}=\frac{z+1}{1}$
• B. $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{1}=\frac{z+\frac{2}{3}}{-1}$
• C. $\frac{x}{0}=\frac{y-1}{-1}=\frac{z+1}{1}$
• D. $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{-1}=\frac{z+\frac{2}{3}}{1}$

Answer 1

The correct option is D $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{-1}=\frac{z+\frac{2}{3}}{1}$

Let $\frac{x-2}{2}=\frac{y-1}{1}=\frac{z}{3}=k$
So, any point on the line is $(2k+2,k+1,3k)$

Let the foot of the perpendicular drawn from the point to the plane is $(x_1,y_1,z_1)$

Now,
$x_1-2(k+1)=y_1-(k+1)=z_1-3k=-\frac{2(k+1)+k+1+3k-1}{1+1+1}$
$\Rightarrow x_1-2(k+1)=y_1-(k+1)=z_1-3k=-(2k+\frac{2}{3})\Rightarrow x_1=\frac{4}{3}+0k;y_1=-k+\frac{1}{3};z_1=k-\frac{2}{3}\Rightarrow k=\frac{x_1-\frac{4}{3}}{0}=\frac{y_1-\frac{1}{3}}{-1}=\frac{z_1+\frac{2}{3}}{1}$

So, the locus will be $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{-1}=\frac{z+\frac{2}{3}}{1}$