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

Question

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

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

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

$\mathrm{R}$ lies on plane $x-y+z=4$

$$
\begin{aligned}
& \therefore \quad 3 \lambda+2-4 \lambda+1+2 \lambda+2=4 \\
& \Rightarrow \quad \lambda=-1 \\
& \therefore \quad R(-1,-5,0)
\end{aligned}
$$

$$
\text { Let } S R \text { be }: \frac{x+1}{2}=\frac{y+5}{2}=\frac{z - 0}{1}=k
$$

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

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

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

$$
\therefore \mathrm{S}=(1,-3,1)
$$

$$
\therefore S R=\sqrt{4+4+1}=3
$$

Solutions · 1