Question

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}$ and $\overrightarrow{c}=\hat{j}-\hat{k}$ be three vectors such that $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ and $\overrightarrow{a}\cdot \overrightarrow{b}=1$. If the length of projection vector of the vector$\overrightarrow{b}$ on the vector $\overrightarrow{a}\times \overrightarrow{c}$ is $l$, then the value of $3l^2$ is equal to

Answer 1

Given: $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$
$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}=|\overrightarrow{c}{|}^2=2$
Also, $\overrightarrow{a}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 0& 1& -1\end{array}\right|=-2\hat{i}+\hat{j}+\hat{k}\Rightarrow |\overrightarrow{a}\times \overrightarrow{c}|=\sqrt{6}$
Now required length of projection
$l=\left|\frac{\overrightarrow{b}\cdot (\overrightarrow{a}\times \overrightarrow{c})}{|\overrightarrow{a}\times \overrightarrow{c}|}\right|=\frac{2}{\sqrt{6}}$
Hence, $3l^2=2$