3D Geometry 3 Question 4
####4. If the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}$ intersects the plane $2 x+3 y-z+13=0$ at a point $P$ and the plane $3 x+y+4 z=16$ at a point $Q$, then $P Q$ is equal to
(2019 Main, 12 April I)
(a) 14
(b) $\sqrt{14}$
(c) $2 \sqrt{7}$
(d) $2 \sqrt{14}$
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Answer:
(d)
Solution:
- Equation of given line is
$ \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=r \text { (let) } $
Now, coordinates of a general point over given
line is $R(3 r+2,2 r-1,-r+1)$
Let the coordinates of point $P$ are $\left(3 r_{1}+2,2 r_{1}-1, r_{1}+1\right)$ and $Q$ are $\left(3 r_{2}+2,2 r_{2}-1,-r_{2}+1\right)$.
Since, $P$ is the point of intersection of line (i) and the plane $2 x+3 y-z+13=0$, so
$ \begin{aligned} & 2\left(3 r_{1}+2\right)+3\left(2 r_{1}-1\right)-\left(-r_{1}+1\right)+13=0 \\ \Rightarrow \quad & 6 r_{1}+4+6 r_{1}-3+r_{1}-1+13=0 \\ \Rightarrow \quad & 13 r_{1}+13=0 \Rightarrow r_{1}=-1 \end{aligned} $
So, point $P(-1,-3,2)$
And, similarly for point ’ $Q$ ‘, we get
$ \begin{aligned} & 3\left(3 r_{2}+2\right)+\left(2 r_{2}-1\right)+4\left(-r_{2}+1\right)=16 \\ \Rightarrow \quad & 7 r_{2}=7 \Rightarrow r_{2}=1 \end{aligned} $
So, point is $Q(5,1,0)$
$ \text { Now, } \quad \begin{aligned} P Q & =\sqrt{(5+1)^{2}+(1+3)^{2}+2^{2}} \\ & =\sqrt{36+16+4} \\ & =\sqrt{56}=2 \sqrt{14} \end{aligned} $
