3D Geometry 3 Question 10
####10. If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane, $x+2 y+3 z=15$ at a point $P$, then the distance of $P$ from the origin is
(2019 Main, 9 April I)
(a) $7 / 2$
(b) $9 / 2$
(c) $\sqrt{5} / 2$
(d) $2 \sqrt{5}$
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Answer:
(b)
Solution:
- Equation of given plane is
$x+2 y+3 z=15$ and line is, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=r$ (let)
So, the coordinates of any point on line (ii) is
$P(1+2 r,-1+3 r, 2+4 r)$.
$\because$ Point $P$ is intersecting point of plane (i) and line (ii)
$\therefore(1+2 r)+2(-1+3 r)+3(2+4 r)=15$
$\Rightarrow 1+2 r-2+6 r+6+12 r=15 \Rightarrow 20 r=10$
$\Rightarrow r=\frac{1}{2}$
$\therefore$ Coordinates of $P=1+1,-1+\frac{3}{2}, 2+2=2, \frac{1}{2}, 4$
Now, distance of the point $P$ from the origin
$=\sqrt{4+\frac{1}{4}+16}=\sqrt{20+\frac{1}{4}}=\sqrt{\frac{81}{4}}=\frac{9}{2}$ units
