The co-ordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meet the plane $2…
The co-ordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meet the plane
$2 x+4 y-z=1$ are
$(3,-1,-1)$
$(3,-1,1)$
$(3,1,-1)$
$(-2,1,-1)$
Solution
Let $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=\lambda$ and $P$ be any point on the given line.
$\therefore P=(2 \lambda+1,-3 \lambda+2,4 \lambda-3)$
Since point $P$ lies on the plane, we write
$2(2 \lambda+1)+4(-3 \lambda+2)-(4 \lambda-3)=1$
$4 \lambda+2+8-12 \lambda-4 \lambda+3=1 \Rightarrow \lambda=1$
$\therefore \mathrm{P} \equiv(3,-1,1)$