Three Dimensional Geometry Question 55
Question: If a,b,g be the direction angles of a vector and $ \cos \alpha =\frac{14}{15} $ , $ \cos \beta =\frac{1}{3} $ then $ \cos \gamma $ =
Options:
A) $ \pm \frac{2}{15} $
B) $ \frac{1}{5} $
C) $ \pm \frac{1}{15} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1 $
$ \Rightarrow \cos ,\gamma =\sqrt{1-{{( \frac{14}{15} )}^{2}}-{{( \frac{1}{3} )}^{2}}} $ $ =\sqrt{\frac{8}{9}-( \frac{196}{225} )}=\pm \frac{2}{15} $ .
BETA
