3D Geometry 3 Question 37
####37. The distance of the point $(1,0,2)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$ is
(2015 Main)
(a) $2 \sqrt{14}$
(b) 8
(c) $3 \sqrt{21}$
(d) 13
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Answer:
Correct Answer: 37. (d)
Solution:
- Given equation of line is
$ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda $
and equation of plane is
$ x-y+z=16 $
Any point on the line (i) is, $(3 \lambda+2,4 \lambda-1,12 \lambda+2)$
Let this point of intersection of the line and plane.
$ \begin{array}{rlrl} \therefore & (3 \lambda+2)-(4 \lambda-1)+(12 \lambda+2) & =16 \\ \Rightarrow & 11 \lambda+5 =16 \\ \Rightarrow & 11 \lambda =11 \\ \Rightarrow & \lambda =1 \end{array} $
So, the point of intersection is $(5,3,14)$.
Now, distance between the points $(1,0,2)$ and $(5,3,14)$
$ \begin{aligned} & =\sqrt{(5-1)^{2}+(3-0)^{2}+(14-2)^{2}} \\ & =\sqrt{16+9+144}=\sqrt{169}=13 \end{aligned} $
