If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}

Question

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} 	imes \vec{b})-3 \vec{b}$ then angle between $\vec{b}$ and $\vec{c}$ is :

Professor PC Thomas Classes · Answer

Given $\vec{c}=(2 \vec{a} 	imes \vec{b})-3 \vec{b}$
$$
\begin{aligned}
& \vec{b} \cdot \vec{c}=\vec{b} \cdot\{(2 \vec{a} 	imes \vec{b})-3 \vec{b}\} \
& \vec{b} \cdot \vec{c}=-3|\vec{b}|^2 \quad \ldots .(1)
\end{aligned}
$$
$$
\begin{aligned}
& 	ext { now }|\vec{a} 	imes \vec{b}|^2=\vec{a}^2 \vec{b}^2-(\vec{a} \cdot \vec{b})^2 \
& =16-4=12 \
& 	ext { and } \vec{c}^2=((2 \vec{a} 	imes \vec{b})-3 \vec{b})^2 \
& |\vec{c}|^2=4|\vec{a} 	imes \vec{b}|^2+9 \vec{b}^2-\underbrace{6 \vec{b} \cdot(2 \vec{a} 	imes \vec{b})}_{	ext {zero }} \
& =48+144=192 \
& |\vec{c}|^2=8 \sqrt{3}
\end{aligned}
$$
Now, $\cos 	heta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}=\frac{-3|\vec{b}|^2}{|\vec{b}||\vec{c}|}=\frac{-3|\vec{b}|}{|\vec{c}|}=$
$$
\frac{-3(4)}{8 \sqrt{3}}=\frac{-\sqrt{ } 3}{2}
$$

Hence $\overrightarrow{\boldsymbol{b}} \wedge \overrightarrow{\boldsymbol{c}}=\frac{5 \pi}{6}$ Ans.