# Prove the following

Question:

Let $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}+5 \hat{k}$. If $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{r}}, \quad \overrightarrow{\mathrm{r}} \cdot(\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$ 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 $\alpha+|\overrightarrow{\mathrm{r}}|^{2}$ is equal to :

1. 9

2. 15

3. 13

4. 11

Correct Option: , 2

Solution:

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

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

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

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

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

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

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

Solve (2)\&(3)

$\alpha=1, \lambda=1$

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

$|\overrightarrow{\mathrm{r}}|^{2}=14 \& \alpha=1$

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