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Question

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 $$\theta$$.


Solution

Given vectors are $$\vec{a}=\hat{i} -2\hat{j}+3\hat{k}$$ and $$\vec{b}=3\hat{i}-2\hat{j}+\hat{k}$$,

If $$\theta$$ is the angle between the given vectors, then 
$$\cos \theta=\dfrac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$$

Now, $$\vec{a}.\vec{b}=$$ $$(\hat{i} -2\hat{j}+3\hat{k}).$$$$(3\hat{i}-2\hat{j}+\hat{k})$$
$$=3+4+3=10$$

Also, $$|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}$$ and 

$$|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1}=\sqrt{14}$$

$$\therefore \cos \theta=\dfrac{10}{14}=\dfrac{5}{7}$$

$$\Rightarrow \theta=\cos^{-1}\left(\dfrac{5}{7}\right)$$

Hence, $$\theta=\cos^{-1}\left(\dfrac{5}{7}\right)$$ is the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$.

Mathematics

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