Three Dimensional Geometry Question 398
Question: The distance between two points P and Q is d and the length of their projections of PQ on the co-ordinate planes are $ d_1,d_2,d_3 $ . Then $ d_1^{2}+d_2^{2}+d_3^{2}=kd^{2} $ where . k- is
Options:
A) 1
B) 5
C) 3
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
Here,
$ d_1=d\cos (90^{o}-\alpha ) $
$ d_2=d\cos (90^{o}-\beta ) $
$ d_3=d\cos (90^{o}-\gamma ) $
Þ $ d_1^{2}+d_2^{2}+d_3^{2}=d^{2}({{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma ) $
Þ $ d_1^{2}+d_2^{2}+d_3^{2}=2d^{2} $ ; \ $ k=2 $ .
BETA
