Mathematics - Vector Question with Solution | TestHub

Given: ${\left|\overrightarrow{b}\right|}^2=6, \overrightarrow{a}·\overrightarrow{b}=3\sqrt{2}, \overrightarrow{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$

We know that, $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$

$\Rightarrow 3\sqrt{2}=\sqrt{1^2+{\alpha }^2+{\beta }^2}\left(\sqrt{6}\right)\cos \frac{\pi }{4}$

$\Rightarrow 3\sqrt{2}=\sqrt{1^2+{\alpha }^2+{\beta }^2}\left(\sqrt{3}\right)$

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

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

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

Also, $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta$

$\Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{6}\sqrt{6}\sin \frac{\pi }{4}=\frac{6}{\sqrt{2}}$

$\Rightarrow {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=18$

$\Rightarrow \left({\alpha }^2+{\beta }^2\right){\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=5\times 18$

$\Rightarrow \left({\alpha }^2+{\beta }^2\right){\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=90$