Three Dimensional Geometry Question 108
Question: The equation of the straight line passing through (1, 2, 3) and perpendicular to the plane $ x+2y-5z+9=0 $ is
[MP PET 1991]
Options:
A) $ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5} $
B) $ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+5}{3} $
C) $ \frac{x+1}{1}=\frac{y+2}{2}=\frac{z+3}{-5} $
D) $ \frac{x+1}{1}=\frac{y+2}{2}=\frac{z-5}{3} $
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Answer:
Correct Answer: A
Solution:
The line passes through point (1, 2, 3) is $ \frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c} $ and it is perpendicular to the plane $ x+2y-5z+9=0, $ therefore the line must be $ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5} $ because $ \sin \theta =\frac{1.1+2.2+(-,5)(-,5)}{\sqrt{1^{2}+2^{2}+5^{2}},.\sqrt{1^{2}+2^{2}+5^{2}}}=1 $ $ ,\Rightarrow \theta =90^{o} $ .
BETA
