If \theta is the angle between two vectors \hat{i}-2\hat{j}+3\hat{k} and 3\hat{i}-2\hat{j}+\hat{k}, find \sin \theta \cdot

 

 

 

 
 
 
 
 

Answers (1)

Let \vec{a}= \hat{i}-2\hat{j}+3\hat{k}
     \vec{b}= 3\hat{i}-2\hat{j}+\hat{k}
we know \vec{a}\cdot \vec{b}=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta
\Rightarrow \cos \theta = \frac{\vec{a}\cdot \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}
= \frac{\left ( \hat{i}-2\hat{j}+3\hat{k} \right )\cdot \left ( 3\hat{i}-2\hat{j}+\hat{k} \right )}{\left | \hat{i}-2\hat{j}+3\hat{k} \right |\cdot \left | 3\hat{i}-2\hat{j}+\hat{k} \right |}
= \frac{1\left ( 3 \right )+\left ( - 2\right )\left ( -2 \right )+3\left ( 1 \right )}{\sqrt{1+4+9}\sqrt{9+4+1}}
=\frac{3+4+3}{14}= \frac{10}{14}= \frac{5}{7}
Now, \sin \theta = \sqrt{1-\cos ^{2}\theta }
                 = \sqrt{1-\left ( \frac{5}{7} \right )^{2}}= \sqrt{1-\frac{25}{49}}
                \Rightarrow \sqrt{\frac{49-25}{49}}\Rightarrow \sqrt{\frac{24}{49}}
\sin \theta = \sqrt{\frac{24}{49}}= \frac{2\sqrt{6}}{7}

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