Three Dimensional Geometry Question 274
Question: The equation of the perpendicular from the point $ (\alpha ,\beta ,\gamma ) $ to the plane $ ax+by+cz+d=0 $ is
[MP PET 2003]
Options:
A) $ a(x-\alpha )+b(y-\beta )+c(z-\gamma )=0 $
B) $ \frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c} $
C) $ a(x-\alpha )+b(y-\beta )+c(z-\gamma )=abc $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
It is a line passing through $ (\alpha ,\beta ,\gamma ) $ and whose direction cosines are $ a,b,c. $
BETA
