Question

If $\overrightarrow{A}=(1,1,1),\overrightarrow{C}=(0,1,-1)$ are given a vector $\overrightarrow{B}$ satisfying the equation $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{C}$ and $\overrightarrow{A}\cdot \overrightarrow{B}=3$ is ______.

Answer 1

Given:$\overrightarrow{A}=(1,1,1),\overrightarrow{C}=(0,1,-1)$

$\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{C}$ & $\overrightarrow{A}.\overrightarrow{B}=3$

To find $\overrightarrow{B}$

Sol Let $\overrightarrow{B}=a\hat{i}+b\hat{j}+c\hat{k}$

$\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{C}$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ a& b& c\end{array}\right|=\hat{j}-\hat{k}$

$(c-b)\hat{i}-\hat{j}(c-a)+\hat{k}(b-a)=\hat{j}-\hat{k}$

$\Rightarrow a-c=1...\left(1\right)$

$-a+b=-1\Rightarrow b=a-1...\left(ii\right)$

$\overrightarrow{A}.\overrightarrow{B}=3$

$(\hat{i}+\hat{j}+\hat{k})(a\hat{i}+b\hat{j}+c\hat{k})=3$

$a+b+c=3...\left(iii\right)$

put (ii) in (iii)

$a+a-1+c=3$

$2a+c=1\left(iv\right)$

Add (1) & (iv)

$a-c=1$

$2a+c=4$

_____________

$3a=5$

$a=5/3,b=2/3,c=2/3$

$\therefore \overrightarrow{B}=(\frac{5}{3},\frac{2}{3},\frac{2}{3})$