Mathematics - Vector Question with Solution | TestHub

Given,

$\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{b}$

$\Rightarrow \overrightarrow{r}\times \overrightarrow{b}-\overrightarrow{c}\times \overrightarrow{b}=0$

$\left(\overrightarrow{r}-\overrightarrow{c}\right)\times \overrightarrow{b}=0$

Hence, we can say that $\overrightarrow{r}-\overrightarrow{c}$ is parallel to vector $\overrightarrow{b}$,

Now let , $\overrightarrow{r}-\overrightarrow{c}=\lambda \overrightarrow{b}$

$\Rightarrow \overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{b} ......\left(1\right)$

$\because \overrightarrow{r}·\overrightarrow{a}=0\Rightarrow \overrightarrow{c}·\overrightarrow{a}+\lambda \left(\overrightarrow{b}·\overrightarrow{a}\right)=0$

$\Rightarrow \lambda =\frac{-\overrightarrow{c}·\overrightarrow{a}}{\overrightarrow{b}·\overrightarrow{a}}=\frac{-8}{5}$

Now putting the value of $\lambda$ in equation $\left(1\right)$ we get,

$\overrightarrow{r}=\overrightarrow{c}-\frac{8}{5}\overrightarrow{b}=\left(5\hat{i}-3\hat{j}+3\hat{k}\right)-\frac{8}{5}\left(\hat{i}-\hat{j}+2\hat{k}\right)$

$\Rightarrow 5\overrightarrow{r}=17\hat{i}-7\hat{j}-\hat{k}$

$\Rightarrow 25{\left|\overrightarrow{r}\right|}^2={\left(17\right)}^2+{\left(-7\right)}^2+{\left(-1\right)}^2$

$\Rightarrow 25{\left|\overrightarrow{r}\right|}^2=289+49+1=339$