Question

The equation of the projection line of the line $\frac{x+1}{2}=\frac{y+1}{-1}=\frac{z+3}{4}$ on the plane $x+2y+z=6$ is

• A. $\frac{x-1}{4}=\frac{y-3}{7}=\frac{z-1}{10}$
• B. $\frac{x-1}{-4}=\frac{y+3}{7}=\frac{z-1}{10}$
• C. $\frac{x-1}{4}=\frac{y-3}{-7}=\frac{z+1}{10}$
• D. $\frac{x+3}{4}=\frac{y-2}{7}=\frac{z-7}{-10}$

Answer 1

Answer: (C)

From the diagram,
Coordinates of the point $A$ are $\left(2\lambda -1,-\lambda -1,4\lambda -3\right)$ (from the line)
It must satisfy the equation of the plane
$2\lambda -1-2\lambda -2+4\lambda -3=6\Rightarrow \lambda =3$
So, coordinates of the point $A$ are $\left(5,-4,9\right)$
Point $B^{{}^{\prime }}$ must be the foot of perpendicular for point $B$
$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+3}{1}=-\frac{\left(-1-2-3-6\right)}{1+4+1}=2$
$x=1,y=3,z=-1$
Direction ratios of $<A{B}^{{}^{\prime }}>=<4,-7,10>$
So, the equation of projection of line is $\frac{x-1}{4}=\frac{y-3}{-7}=\frac{z+1}{10}$