If $\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$, then the angle between the vectors $(2 \bar{a}+\bar{b})$ and $(\bar{a}+2 \bar{b})$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } 2 \overline{\mathrm{a}}+\overline{\mathrm{b}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}} \ & \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}=5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ & |2 \overline{\mathrm{a}}+\overline{\mathrm{b}}|=\sqrt{(4)^2+(-1)^2+(5)^2}=\sqrt{42} \ & |\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}|=\sqrt{5^2+4^2+1^2}=\sqrt{42} \end{aligned}$ $	herefore \quad$ Angle between $(2 \bar{a}+\bar{b})$ and $(\bar{a}+2 \bar{b})$ is given by $\begin{aligned} \cos 	heta & =\frac{(2 \bar{a}+\bar{b})(\bar{a}+2 \bar{b})}{|2 \bar{a}+\bar{b}||\bar{a}+2 \bar{b}|} \ & =\frac{(4 \hat{i}-\hat{j}+5 \hat{k})(5 \hat{i}+4 \hat{j}+\hat{k})}{\sqrt{42} \cdot \sqrt{42}}=\frac{21}{42} \ \cos 	heta & =\frac{1}{2} \ \Rightarrow 	heta & =\frac{\pi}{3} \end{aligned}$

If $\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$, then the angle between…

Solution