Question

The foot of the perpendicular from (2, 4, -1) to the line $x+5=\frac{1}{4}\left(y+3\right)=-\frac{1}{9}\left(z-6\right)$

• A. (- 4, 1, - 3)
• B. (4, -1, -3)
• C. (-4, -1, 3)
• D. (- 4, - 1, -3)

Answer 1

Answer: (A)

Given equation of line is
$x+5=\frac{1}{4}\left(y+3\right)=-\frac{1}{9}\left(z-6\right)$
or $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda \left(say\right)$
$x=\lambda -5,y=4\lambda -3,z=-9\lambda +6$
$\left(x,y,z\right)\equiv \left(\lambda -5,4\lambda -3,-9\lambda +6\right)\dots \left(i\right)$
Let it is foot of perpendicualr
So, d.r.’s of $\perp$ line is
$\left(\lambda -5-2,4\lambda -3-4,-9\lambda +6+1\right)$
$\equiv \left(\lambda -7,4\lambda -7,-9\lambda +7\right)$
D.r.’s of given line is $\left(1,4,-9\right)$ and both lines are $\perp$
$\therefore \left(\lambda -7\right).1+\left(4\lambda -7\right).4+\left(-9\lambda +7\right)\left(-9\right)=0$
$\Rightarrow 98\lambda =98\Rightarrow \lambda =1$
$\therefore$ Point is $(-4,1,-3)$. [Substituting $\lambda =1$ in (i)]