# Prove the following

Question:

Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=7 \hat{\mathrm{i}}+\hat{\mathrm{j}}-6 \hat{\mathrm{k}}$.

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

is equal to :

1. 12

2. 8

3. 13

4. 10

Correct Option: 1

Solution:

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

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

$\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})$

$\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(-5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})$

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

$\Rightarrow \lambda(-5-8+10)=-3$

$\lambda=1$

Now $\overrightarrow{\mathrm{r}}=-5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$

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

$=-10+12+10=12$