# Solve this following

Question:

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

Solution:

$\vec{a} \times \vec{b}=c$

Take Dot with $\overrightarrow{\mathrm{c}}$

$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|^{2}=2$

Projection of $\vec{b}$ or $\vec{a} \times \vec{c}=\ell$

$\frac{|\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})|}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|}=\ell$

$\therefore \ell=\frac{2}{\sqrt{6}} \Rightarrow \ell^{2}=\frac{4}{6}$

$3 \ell^{2}=2$