# Prove the following

Question:

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be three given vectors. If $\vec{r}$ is a vector such that

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=0$, then $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}$ is equal to

Solution:

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

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=0$

$(\vec{r}-\vec{c}) \times \vec{a}=0$

$\therefore \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}=0$

$\Rightarrow \lambda(1-2)+2=0$

$\Rightarrow \lambda=2$

$\therefore \overrightarrow{\mathrm{r}}=2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=2|\overrightarrow{\mathrm{a}}|^{2}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$

$=2(1+4+1)+(1-2+1)$

$=12$