# let vector a, vector i + vector 2j - vector k,

Question:

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

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

$\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{a}}$

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

$\Rightarrow \lambda=\frac{-\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=-\frac{2}{-1}=2$

So, $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+2 \mathrm{a}^{2}=12$