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Question

Find the cosine of the angle between the vectors $$\vec{a}=i+j+k$$ and $$\vec{b}=2i-3j+2k$$


Solution

We have
$$\vec a\cdot \vec b=ab\cos\theta$$
$$\cos\theta=\left(\dfrac{\vec a\cdot \vec b}{ab}\right)$$
$$\vec a\cdot \vec b=a_xb_x+a_yb_y+a_cb_c$$
        $$=(1)(2)+(1)(-3)+(1)(2)=1$$
$$a=\sqrt{1^2+1^2+1^2}=\sqrt3$$
$$b=\sqrt{2^2+(-3)^2+2^2}=\sqrt17$$
Therefore, we have
$$\cos\theta=\left(\dfrac{1}{\sqrt3\sqrt17}\right)=0.14$$

Maths

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