Let |a|=2,|b|=3 and the angle between the vectors a and b be ( )/(4).

Question

Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi }{4}$. Then $|(\vec{a}+2\vec{b})	imes (2\vec{a}-3\vec{b}){|}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} |\vec{a}|=2 \ |\vec{b}|=3 \ \vec{a}\cdot \vec{b}=\frac{\pi }{4} \end{gathered}$

$\begin{gathered} |(\vec{a}+2\vec{b})	imes (2\vec{a}-3\vec{b}){|}^{2} \ =|-3\vec{a}	imes \vec{b}+4\vec{b}	imes \vec{a}{|}^{2} \ =|-3\vec{a}	imes \vec{b}-4\vec{a}	imes \vec{b}{|}^{2} \ =|-7\vec{a}	imes \vec{b}{|}^{2} \ ={(-7|\vec{a}|	imes |\vec{b}|\sin (\frac{\pi }{4}))}^{2} \end{gathered}$

= $49	imes 4	imes 9	imes \frac{1}{2}=882$

Solutions · 1