Three Dimensional Geometry Question 183
Question: What is the equation of the plane which passes through the z-axis and is perpendicular to the line $ \frac{x-a}{\cos \theta }=\frac{y+2}{\sin \theta }=\frac{z-3}{0} $ ?
Options:
A) $ x+y,tan,\theta =0 $
B) $ y+x,tan,\theta =0 $
C) $ x,\cos \theta -y,\sin \theta =0 $
D) $ x,\sin \theta -y,\cos \theta =0 $
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Answer:
Correct Answer: A
Solution:
[a] The plane is perpendicular to the line $ \frac{x-a}{\cos \theta }=\frac{y+2}{\sin \theta }=\frac{z-3}{0} $ Hence, the direction ratios of the normal of the plane are $ \cos \theta $ , $ \sin \theta $ and 0. …(i) Now, the required plane passes through the z-axis. Hence, the point (0, 0, 0) lies on the plane. From Eqs. (i) and (ii), we get equation of the plane as $ \cos \theta (x-0)+\sin \theta (y-0)+0(z-0)=0 $ $ \cos \theta x+\sin \theta y=0 $ $ x+y\tan \theta =0 $
BETA
