Let a=a_1i+a_2j+a_3k and b=b_1i+b_2j+b_3k be two vectors such that | a|=1,a b=2 and |b|=4.

Question

Let $\overset{	o }{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$ and $\overset{	o }{b}={b}_{1}\hat{i}+{b}_{2}\hat{j}+{b}_{3}\hat{k}$ be two vectors such that $|\overset{	o }{a}|=1,\vec{a}\cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a}	imes \vec{b})-3\vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

Accepted Answer

Given $|\vec{a}|=1,|\vec{b}|=4,\vec{a}\cdot \vec{b}=2$

$\vec{c}=2(\vec{a}	imes \vec{b})-3\vec{b}$

Dot product with $\overset{	o }{a}$ on both sides

$\overset{	o }{c}.\overset{	o }{a}=-6$ ..... (1)

Dot product with $\vec{b}$ on both sides

$\begin{gathered} \overset{	o }{b}\overset{	o }{c}=-48\, \ \overset{	o }{c}\cdot \overset{	o }{c}=4|\overset{	o }{a}	imes \overset{	o }{b}{|}^{2}+9|\overset{	o }{b}{|}^{2} \ |\overset{	o }{c}{|}^{2}=4[|a{|}^{2}|b{|}^{2}-(a\cdot \overset{	o }{b}{)}^{2}]+9|\overset{	o }{b}{|}^{2} \ |\overset{	o }{c}{|}^{2}=4[(1)(4{)}^{2}-(4)]+9(16) \ |\overset{	o }{c}{|}^{2}=4[12]+144 \ |\overset{	o }{c}{|}^{2}=48+144 \ |\overset{	o }{c}{|}^{2}=192 \ 	herefore \cos 	heta =\frac{\overset{	o }{b}\cdot \overset{	o }{c}}{|\overset{	o }{b}|\overset{	o }{c}|} \ 	herefore \cos 	heta =\frac{-48}{\sqrt{192}\cdot 4} \ 	herefore \cos 	heta =\frac{-48}{8\sqrt{3}.4} \ 	herefore \cos 	heta =\frac{-3}{2\sqrt{3}} \ 	herefore \cos 	heta =\frac{-\sqrt{3}}{2}\Rightarrow 	heta ={\cos }^{-1}(\frac{-\sqrt{3}}{2}) \end{gathered}$

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