Question

The point of intersection of the line joining the points $(2,0,2)$ and $(3,-1,3)$ and the plane $x-y+z=1$ is

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

Answer 1

Answer: (A)

$(3,2,0)$

The equation of the line joining the points $(2,0,2)$ & $(3,-1,3)$ is

$\frac{x-2}{3-2}=\frac{y-0}{-1-0}=\frac{z-2}{3-2}\Rightarrow \frac{x-2}{1}=\frac{y-0}{-1-0}=\frac{z-2}{3-2}\Rightarrow \frac{x-2}{1}=\frac{y}{-1}=\frac{z-2}{1}\Rightarrow \frac{x-2}{1}=\frac{y}{-1}=\frac{z-2}{1}=\lambda \left(say\right)\left(\lambda isaparameter\right)\Rightarrow x=\lambda +2,y=-\lambda ,z=\lambda +2$

If , $(\lambda +2,-\lambda ,\lambda +2)$ lies on the plane.

$x-y+z=1$ then

$\lambda +2-(-\lambda )+\lambda +2=1\Rightarrow 2\lambda +\lambda =1-4\Rightarrow 3\lambda =-3\Rightarrow \lambda =-1$

$\therefore$ The point of intersection is $(-1+2,-(-1),-1,2)i.e.(1,1,1)$