Question

Equation of the projection of the line that passes through the two planes $8x-y-7z=8,x+y+z=1$ on the plane $5x-4y-z=5$ is

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

Answer 1

Answer: (A)

$\frac{x-1}{1}=\frac{y}{2}=\frac{z}{-3}$

Any plane through the given line is $x+y+z-1+\lambda (8x-y-7z-8)=0$ ...(1)

$\Rightarrow (1+8\lambda )x+(1-\lambda )y+(1-7\lambda )z-1-8\lambda =0$

It is perpendicular to the plane $5x-4y-z=5$ if $(1+8\lambda )5+(1-\lambda )(-4)-(1-7\lambda )=0\Rightarrow \lambda =0$

So that (1) becomes $x+y+z-1=0$

Now, the line of intersection of the planes $x+y+z-1=0$ and $5x-4y-z=5$ is the required line of projection, which clearly passes through $(1,0,0)$ and if $l,m,n$ are direction radios of line then $l+m+n=0$ and $5l-4m-n=0$

$\Rightarrow \frac{l}{-1+4}=\frac{m}{5+1}=\frac{n}{-4-5}$

$\Rightarrow \frac{l}{3}=\frac{m}{6}=\frac{n}{-9}\Rightarrow \frac{l}{1}=\frac{m}{2}=\frac{n}{-3}$

and hence the required equation of the projection is

$\frac{x-1}{1}=\frac{y}{2}=\frac{z}{-3}$