Question

The point of intersection of the straight line $\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+2}{1}$ with the plane $x+3y-z+1=0$ is

• A. $(3,-1,1)$
• B. $(-5,1,-1)$
• C. $(2,0,3)$
• D. $(5,-1,3)$
• E. $(4,-2,-1)$

Answer 1

Answer: (E)

$(4,-2,-1)$

Given equation of straight line is,
$\frac{x-2}{2}=\frac{y-1}{-3},\frac{z+2}{1}=r$ (say)
So, any point on the line is $(2r+2,-3r+1,r-2)$ which lie on the plane $x+3y-z+1=0$
i.e., $(2r+2)+3(-3r+1)-(r-2)+1=0$
$\Rightarrow 2r+2-9r+3-r+2+1=0$
$\Rightarrow -8r+8=0\Rightarrow r=1$
So, required intersection point is $(4,-2,-1)$.