Question

If $\overrightarrow{A}=3i+j+2k$ and $\overrightarrow{B}=2i-2j+4k$ and $\theta$ is the angle between the two vectors, there $sin\theta$ is equal to

• A. $2/3$
• B. $2/\sqrt{3}$
• C. $2/\sqrt{7}$
• D. $2/\sqrt{1}3$

Answer 1

Answer: (C)

$2/\sqrt{7}$

$Given:$ $\overrightarrow{A}=3i+j+2k$ And $\overrightarrow{B}=2i-2j+4k$

$Solution:$ We know that angle between two vector (a) and (b) is given by

$\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

$=\frac{(3\hat{i}+\hat{j}+2\hat{k})\cdot (2\hat{i}-2\hat{j}+4\hat{k})}{\sqrt{3^2+1^2+2^2}\sqrt{2^2+(-2{)}^2+4^2}}$

$cos\theta =\frac{3}{\sqrt{2}1}$

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{9}{21}}=\frac{2}{\sqrt{7}}$

$So,the$ $correct$ $option:C$