Three Dimensional Geometry Question 351
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-1}{-6} $ is
Options:
A) 1
B) 2
C) 4
D) $ 2\sqrt{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Equation of the line through (1, -2, 3) parallel to the line $ \frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6} $ is $ \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{-6}=r(say) $ ?..(1) Then any point on (1) is $ (2r+1,3r-2,-6r+3) $ if this point lies on the plane $ x-y+z=5 $ then $ (2r+1)-(3r-2)+(-6r+3)=5\Rightarrow r=\frac{1}{7} $ Hence the point is $ ( \frac{9}{7},-\frac{11}{7},\frac{15}{7} ) $ Distance between $ (1,-2,3) $ and $ ( \frac{9}{7},-\frac{11}{7},\frac{15}{7} ) $ $ =\sqrt{( \frac{4}{49}+\frac{9}{49}+\frac{36}{49} )}=\sqrt{( \frac{49}{59} )}=1 $
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