Vector Algebra Question 347
Question: The angle between the vectors $ 3,\mathbf{i}+\mathbf{j}+2,\mathbf{k} $ and $ 2,\mathbf{i}-2,\mathbf{j}+4,\mathbf{k} $ is
[MP PET 1990]
Options:
A) $ {{\cos }^{-1}}\frac{2}{\sqrt{7}} $
B) $ {{\sin }^{-1}}\frac{2}{\sqrt{7}} $
C) $ {{\cos }^{-1}}\frac{2}{\sqrt{5}} $
D) $ {{\sin }^{-1}}\frac{2}{\sqrt{5}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \cos \theta =\frac{3(2)+(1)(-2)+2(4)}{\sqrt{9+1+4}\sqrt{4+4+16}} $ $ =\frac{12}{\sqrt{14}\sqrt{24}}=\frac{6}{\sqrt{14}\sqrt{6}} $
$ \Rightarrow \cos \theta =\frac{\sqrt{3}}{\sqrt{7}}\Rightarrow \sin \theta =\frac{2}{\sqrt{7}} $
Þ $ \theta ={{\sin }^{-1}}( \frac{2}{\sqrt{7}} ) $ .
BETA
