Let a=i+ j+ k, , R.

Question

Let $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k},\alpha ,\beta \in R$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi }{4}$ and $|\vec{b}{|}^{2}=6$. If $\vec{a}\cdot \vec{b}=3\sqrt{2}$, then the value of $({\alpha }^{2}+{\beta }^{2})|\vec{a}	imes \vec{b}{|}^{2}$ is equal to

Accepted Answer

$\begin{gathered} |\overset{	o }{b}{|}^{2}=6;|\overset{	o }{a}||\overset{	o }{b}|\cos 	heta =3\sqrt{2} \ |\overset{	o }{a}{|}^{2}|\overset{	o }{b}{|}^{2}{\cos }^{2}	heta =18 \ |\overset{	o }{a}{|}^{2}=6 \end{gathered}$

Also $1+{\alpha }^{2}+{\beta }^{2}=6$

${\alpha }^{2}+{\beta }^{2}=5$

to find

$\begin{gathered} ({\alpha }^{2}+{\beta }^{2})|\vec{a}{|}^{2}|\vec{b}{|}^{2}{\sin }^{2}	heta \ =(5)(6)(6)(\frac{1}{2}) \ =90 \end{gathered}$

Solutions · 1