Let $\vec{a} \times \vec{b}=7 \hat{i}-5 \hat{j}-4 \hat{k}$ and $\vec{a}=\hat{i}+3 \hat{j}-2 \hat{k}$. If the…
Let $\vec{a} \times \vec{b}=7 \hat{i}-5 \hat{j}-4 \hat{k}$ and $\vec{a}=\hat{i}+3 \hat{j}-2 \hat{k}$. If the length of projection of $\vec{b}$ on $\vec{a}$ is $\frac{8}{\sqrt{14}}$, then $|\vec{b}|=$
$121$
$\sqrt{12}$
$\sqrt{11}$
$144$
Solution
Since, length of projection of $\vec{b}$ on $\vec{a}$ is $\frac{8}{\sqrt{14}}$
$\Rightarrow \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}|}=\frac{8}{\sqrt{14}}$
Now, $\vec{a} \times \vec{b}=\frac{\vec{a} \times \vec{b}}{|\vec{a}|}=\frac{7 \hat{i}-5 \hat{j}-4 \hat{k}}{\sqrt{1+9+4}}=\frac{7 \hat{i}-5 \hat{j}-4 \hat{k}}{\sqrt{14}}$
Since, $|\vec{a} \cdot \vec{b}|^2+|\vec{a} \times \vec{b}|^2=|\vec{b}|^2|\vec{a}|^2$
$\Rightarrow|\vec{b}|^2=\frac{154}{14} \Rightarrow|\vec{b}|=\sqrt{11}$