Question

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

• A. $2x+y+2=0$
• B. $3x+y-z=2$
• C. $2x-3y+8z=3$
• D. $4x+3y+4=0$

Answer 1

The correct option is A $2x+y+2=0$

Given line, $\frac{x+1}{-1}=\frac{y}{2}=\frac{z-1}{3}$
passing through $A(-1,0,1)$ and $d.r^{\prime }s(a_1,b_1,c_1)=(-1,2,3)$
and the plane $x-2y+z=6$ is with normal $d.r^{\prime }s(a_2,b_2,c_2)=(1,-2,1)$
So,the projection plane is :
$\left|\begin{array}{ccc}x+1& y& z-1\\ -1& 2& 3\\ 1& -2& 1\end{array}\right|=0\Rightarrow 2x+y+2=0$