Question

If $\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{b}\&\overrightarrow{r},\overrightarrow{a}=0$ where
$\overrightarrow{a}=2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}-\stackrel{\wedge }{k},\overrightarrow{b}=3\stackrel{\wedge }{i}-\stackrel{\wedge }{j}-\stackrel{}{k,\overrightarrow{c}=\stackrel{\wedge }{i}+\stackrel{\wedge }{j}+\stackrel{\wedge }{k}}$ then $\overrightarrow{r}$

• A. $2(\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+\stackrel{\wedge }{k)}$
• B. $2(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}-\stackrel{\wedge }{k)}$
• C. $2(-\stackrel{\wedge }{i}+\stackrel{\wedge }{j}+\stackrel{\wedge }{k)}$
• D. $2(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}+\stackrel{\wedge }{k)}$

Answer 1

The correct option is B $2(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}-\stackrel{\wedge }{k)}$

$\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{b};\overrightarrow{r}\times \overrightarrow{a}=0,$

$\text{Let}r=x\hat{i}+y\hat{j}+z\hat{k}$

so $\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ x& y& z\\ 3& -1& -1\end{array}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ 1& 1& 1\\ 3& -1& -1\end{array}\right|$

$\Rightarrow \hat{i}\left(-y+z\right)-\hat{j}\left(-x-3z\right)+\hat{k}\left(-x-3y\right)=\hat{i}\left(0\right)-\hat{j}\left(-4\right)+\hat{k}\left(-4\right)$

$\text{Comparing,we}\text{get}$

$\text{- y + z = 0}\Rightarrow \text{y}\text{=}\text{z}$

$\&-x-3y=-4or-x-3z=-4orx+3y=\mathrm{4...........}\left(1\right)$

$Also\overrightarrow{r.}\overrightarrow{a}=0$

$\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(x\hat{i}+y\hat{j}+z\hat{k}\right)=0$

$2x+3y-z=0$

$x+3y-y=0$ $(asy=z)$

$2x=-2y$

$x=-y$

put in ........(1)

2y = 4

Y = 2

$\therefore$ X= - 2

Z=2

$\overrightarrow{r}=-2\hat{i}+2\hat{j}+2\hat{k}=2\left(-\hat{i}+\hat{j}+\hat{k}\right)$