Question

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of perpendiculars lie on the line is

• A. $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$
• B. $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$
• C. $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$
• D. $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

Answer 1

The correct option is D $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

Equation of the given line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$

So any point on line is $x=2\lambda -2,y=-\lambda -1,z=3\lambda$

If it lies on the plane $x+y+z=3$, then $(2\lambda -2)+(-\lambda -1)+\left(3\lambda \right)=3$....$\left(1\right)$

$\Rightarrow 4\lambda =6\Rightarrow \lambda =\frac{3}{2}$....$\left(2\right)$

$\Rightarrow$ the point of intersection of the line and the plane has coordinates $A$ $(1,\frac{-5}{2},\frac{9}{2})$.....$\left(3\right)$

Now, from the equation of the line, we observe that $(-2,-1,0)$ is a point on it.

Let $(x,y,z)$ be the foot of the perpendicular from point $(-2,-1,0)$ on the plane $x+y+z=3$.

$\Rightarrow \frac{x+2}{1}=\frac{y+1}{1}=\frac{z-0}{1}=\frac{-\left[1\right(-2)+1(-1)+0\times 1-3]}{1^2+1^2+1^2}$

$\Rightarrow B(x,y,z)\equiv (0,1,2)$....$\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$ direction ratios of $AB=\left(\right(1,\frac{-7}{2},\frac{5}{2})\equiv (2,-7,5\left)\right)$....$\left(5\right)$

Hence, from $\left(4\right)$ and $\left(5\right)$, equation of the required line is $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$