Three Dimensional Geometry Question 162
Question: The distance of the point (?1, ?5, ?10) 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=5 $ , is
[AISSE 1985; DSSE 1984; MP PET 2002]
Options:
A) 10
B) 11
C) 12
D) 13
Show Answer
Answer:
Correct Answer: D
Solution:
Any point on the line $ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t $ is $ (3t+2,,4t-1,,12t+2) $ This lies on $ x-y+z=5 $
$ \therefore $ $ 3t+2-4t+1+12t+2=5 $ i.e., $ 11t=0\Rightarrow t=0 $
$ \therefore $ Point is $ (2,,-1,,2) $ . Its distance from $ (-1,,-5,,-10) $ is, = $ \sqrt{{{(2+1)}^{2}}+{{(-1+5)}^{2}}+{{(2+10)}^{2}}} $ = $ \sqrt{9+16+144}=13 $ .
BETA
