Three Dimensional Geometry Question 403
Question: If O is the origin and A is the point (a, b, c) then the equation of the plane through A and at right angles to OA is
[AMU 2005]
Options:
A) $ a(x-a)-b(y-b)-c(z-c)=0 $
B) $ a(x+a)+b(y+b)+c(z+c)=0 $
C) $ a(x-a)+b(y-b)+c(z-c)=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Normal will be OA whose direction ratios are $ a-0, $ $ b-0, $ $ c-0 $ i.e., a, b, c. It passes through $ A,(a,b,c). $
$ \therefore $ Equation of required plane is, $ a,(x-a)+b,(y-b)+c,(z-c)=0 $ .
BETA
