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Question

Find the angle between the vectors $$ \left (2\hat{i}+\hat{j}+3\hat{k} \right )$$ and $$ \left (3\hat{i}-2\hat{j}+\hat{k} \right )$$.


Solution

Let $$\vec{a}=2\hat{i}+\hat{j}+3\hat{k}, \vec{b}=3\hat{i}-2\hat{j}+\hat{k}$$

We know that $$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|cos\:\theta$$, where $$\theta$$ is the angle between $$a$$ and $$b$$.

$$\vec{a} \cdot \vec{b}=(2\hat{i}+\hat{j}+3\hat{k}) \cdot (3\hat{i}-2\hat{j}+\hat{k})$$

        $$=6-2+3$$

$$\vec{a} \cdot \vec{b}=7$$

$$|\vec{a}|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$$

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

Substitute the values in $$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|cos\:\theta$$ we get

$$7=\sqrt{14}\cdot \sqrt{14} cos\:\theta$$

$$\Rightarrow 7=14cos\:\theta$$

$$\Rightarrow cos\: \theta=\dfrac{7}{14}$$

$$\Rightarrow cos\: \theta=\dfrac{1}{2}$$

$$\Rightarrow \theta = cos^{-1} \left(\dfrac{1}{2}\right)$$

$$\Rightarrow \theta=\dfrac{\pi}{3}$$

Mathematics

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