Let the line passing through the points P(2,-1,2) and Q(5,3,4) meet the plane x-y+z=4 at t…

Question

Let the line passing through the points $P(2,-1,2)$ and $Q(5,3,4)$ meet the plane $x-y+z=4$ at the point $R$. Then the distance of the point $R$ from the plane $x+2y+3z+2=0$ measured parallel to the line $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is equal to :

Accepted Answer

Equation of line $PQ$ :

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

Let $R$ be $(3\lambda +2,4\lambda -1,2\lambda +2)$

$R$ lies on plane $x-y+z=4$

$\begin{gathered} 	herefore \, 3\lambda +2-4\lambda +1+2\lambda +2=4 \ \Rightarrow \, \lambda =-1 \ 	herefore \, R(-1,-5,0) \end{gathered}$

$SR:\frac{x+1}{2}=\frac{y+5}{2}=\frac{z-0}{1}=k$

Point $S:(2k-1,2k-5,k)$

$S$ lies on plane: $x+2y+3z+2=0$

$\begin{gathered} \Rightarrow (2k-1)+(4k-10)+3k+2=0 \ \Rightarrow 9k-9=0\Rightarrow k=1 \end{gathered}$

$	herefore S=(1,-3,1)$

$	herefore SR=\sqrt{4+4+1}=3$

Solutions · 1