Question

The projection of the line $\frac{x-1}{7}=\frac{y-2}{-4}=\frac{z+3}{-3}$ in the plane $3x-3y+10z-26=0,$ is

• A. $\frac{(x-5)}{827}=\frac{y+2}{-481}=\frac{2z-3}{-113}$
• B. $\frac{2x-5}{817}=\frac{2y-1}{-463}=\frac{z-2}{-192}$
• C. $\frac{2x+1}{829}=\frac{2y-7}{-481}=\frac{2z+8}{-162}$
• D. $\frac{x+1}{825}=\frac{y-7}{-481}=\frac{z+8}{-162}$

Answer 1

The correct option is B $\frac{2x-5}{817}=\frac{2y-1}{-463}=\frac{z-2}{-192}$

Given ,
line$\frac{x-1}{7}=\frac{y-2}{-4}=\frac{z+3}{-3}$ passes through $(1,2,-3)$ and $d.r^{\prime }s=(7,-4,-3)$
given plane,
$3x-3y+10z-26=0,$
let $M(p,q,r)$ is the foot of perpendicular of point $(1,2,-3)$in the plane is given by,
$\frac{p-1}{3}=\frac{q-2}{-3}=\frac{r+3}{10}=-\frac{\left(3\right(1)-3(2)+10(-3)-26)}{3^2+3^2+{10}^2}=\frac{1}{2}$
on solving we get,
$M(p,q,r)=(\frac{5}{2},\frac{1}{2},2)$
Now consider $N$ is point of intersection of line and plane.
Variable point on line is $(7k+1,-4k+2,-3k-3)$ lies on $3x-3y+10z-26=0\Rightarrow N=(\frac{416}{3},-\frac{230}{3},-62)$
$\therefore$D'R's of $MN=(817,-463,-384)$
So projection line equation is $\frac{x-\frac{5}{2}}{817}=\frac{y-\frac{1}{2}}{-463}=\frac{z-2}{-384}$
$\Rightarrow \frac{2x-5}{817}=\frac{2y-1}{-463}=\frac{z-2}{-192}$