If a and b are two vectors such that |a|=2 and |b|=3, then the maximum value of 3|(3a+2b)|…

Question

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$ and $|\vec{b}|=3$, then the maximum value of $3|(3\vec{a}+2\vec{b})|+4|(3\vec{a}-2\vec{b})|$ is :

Accepted Answer

$\begin{gathered} E=3\sqrt{9{a}^{2}+4{b}^{2}+12\vec{a}\cdot \vec{b}}+4\sqrt{9{a}^{2}+4{b}^{2}-12\vec{a}\cdot \vec{b}} \ =3\sqrt{36+36+12	imes 6\cos 	heta }+4\sqrt{36+36-72\cos 	heta } \ =3\sqrt{72+72\cos 	heta }+4\sqrt{72-72\cos 	heta } \ =18\sqrt{2}\sqrt{1+\cos 	heta }+24\sqrt{2}\sqrt{1-\cos 	heta } \ =18\sqrt{2}\sqrt{2}\cos \frac{	heta }{2}+24\sqrt{2}(\sqrt{2}\sin \frac{	heta }{2}) \ =36(\cos \frac{	heta }{2})+48\sin (\frac{	heta }{2}) \ {E}_{max}=\sqrt{(36{)}^{2}+(48{)}^{2}}=\sqrt{{6}^{4}+{6}^{2}	imes {8}^{2}}=60 \end{gathered}$

Solutions · 1