Let a=i+2j+3k,b=i-j+2k and c=5i-3j+3k be three vectors.

Question

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=5\hat{i}-3\hat{j}+3\hat{k}$ be three vectors. If $\vec{r}$ is a vector such

that, $\vec{r}	imes \vec{b}=\vec{c}	imes \vec{b}$ and $\vec{r}\cdot \vec{a}=0$, then $25|\vec{r}{|}^{2}$ is equal to :

Accepted Answer

$\overset{	o }{a}=\hat{i}+2\hat{j}+3\hat{k}$

$\overset{	o }{b}=\hat{i}-\hat{j}+2\hat{k}$

$\overset{	o }{c}=\hat{5i}-3\hat{j}+3\hat{k}$

$(\overset{	o }{r}-\overset{	o }{c})	imes \vec{b}=0,\, \vec{r}\cdot \vec{a}=0$

$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$

Also, $(\vec{c}+\lambda \vec{b})\cdot \vec{a}=0$

$\Rightarrow \vec{a}\cdot \vec{c}+\lambda (\vec{a}\cdot \vec{b})=0$

$	herefore \lambda =\frac{\vec{a}\cdot \vec{c}}{\vec{a}\cdot \vec{b}}=\frac{-8}{5}$

$\overset{	o }{r}=\frac{5(5\hat{i}-3\hat{i}+3\hat{k})-8(\hat{i}-\hat{j}+2\hat{k})}{5}$

$\Rightarrow$ $\overset{	o }{r}=\frac{17\hat{i}-7\hat{j}+\hat{k}}{5}$

$\Rightarrow$ $|\overset{	o }{r}{|}^{2}=\frac{1}{25}(289+50)$

$\Rightarrow$ $25|\overset{	o }{r}{|}^{2}=339$

Solutions · 1