# Prove the following

Question:

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$. If

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{r}} \cdot(\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$ and

$\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\alpha \hat{\mathrm{k}})=-1, \alpha \in \mathrm{R}$, then the value of $\alpha+|\vec{r}|^{2}$ is equal to :

1. 9

2. 15

3. 13

4. 11

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{r}} \Rightarrow \overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=0$

$\overrightarrow{\mathrm{r}}=\vec{\lambda}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \Rightarrow \overrightarrow{\mathrm{r}}=\vec{\lambda}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$

$\overrightarrow{\mathrm{r}}=\vec{\lambda}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})$.......(1)

$\overrightarrow{\mathrm{r}} \cdot(\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$

Put $\overrightarrow{\mathrm{r}}$ from (1) $\alpha \lambda=1$.......(2)

$\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\alpha \hat{\mathrm{k}})=-1$

Put $\overrightarrow{\mathrm{r}}$ from (1) $2 \lambda \alpha-\lambda=1$......(3)

Solve (2) & (3)

$\alpha=1, \quad \lambda=1$

$\Rightarrow \quad \overrightarrow{\mathrm{r}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$|\vec{r}|^{2}=14 \quad \& \quad \alpha=1$

$\alpha+|\vec{r}|^{2}=15$