Three Dimensional Geometry Question 307
Question: The distance of the point (1, -2, 3) from the plane $ x-y+z=5 $ measured parallel to the line $ \frac{x}{2}=\frac{y}{3}=\frac{z}{-6}, $ is
[AI CBSE 1984]
Options:
A) 1
B) 6/7
C) 7/6
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Direction cosines of line $ =( \frac{2}{7},\frac{3}{7},\frac{-6}{7} ) $ Now, $ {x}’=1+\frac{2r}{7},{y}’=-2+\frac{3r}{7} $ and $ {z}’=3-\frac{6r}{7} $
$ \therefore ,( 1+\frac{2r}{7} )-( -2+\frac{3r}{7} )+( 3-\frac{6r}{7} )=5\Rightarrow r=1. $
BETA
