Question

The line joining the points $(2,1,3)$ and $(4,-2,5)$ cuts the plane $2x+y-z=3$.

Where does the line cut the plane?

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

Answer 1

The correct option is D $(0,4,1)$

Equation of line joining two points A=(2,1,3) and B=(4,-2,5),

$\frac{x-2}{4-2}=\frac{y-1}{-2-1}=\frac{z-3}{5-3}=\lambda$

$\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z-3}{2}=\lambda .............\left(1\right)$

let general point P lies on plane $2x+y-z=3$

$P=(2\lambda +2,-3\lambda +1,2\lambda +3)$

Since point P lies on plane $2x+y-z=3$

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

$4\lambda +4-3\lambda +1-2\lambda -3=3$

$\lambda =-1$

Therefore point P is,

$P=(0,4,1)$

Line (1) cuts plane at point $P(0,4,1)$

option $\left(D\right)$ is correct.