Question

Let $\overline{a}=\overline{i}+\overrightarrow{j}+\overline{k},\overline{c}=\overrightarrow{j}-\overline{k}.$ If $\overline{b}$ is a vector satisfying $\overline{a}\times \overline{b}=\overline{c}$ and $\overline{a}.\overline{b}=3,$ then $\overline{b}.$

• A. $\frac{1}{3}(5\overline{i}+2\overline{j}+2\overline{k})$
• B. $\frac{1}{3}(5\overline{i}-2\overline{j}-2\overline{k})$
• C. $3\overline{i}-\overline{j}-\overline{k}$
• D. $\frac{1}{3}(3\overline{i}-\overline{j}-\overline{k})$

Answer 1

The correct option is A $\frac{1}{3}(5\overline{i}+2\overline{j}+2\overline{k})$

$\overrightarrow{a}=\hat{ı}+\hat{ȷ}+\hat{k}\overrightarrow{c}=\hat{ȷ}-\hat{k}$

Let $\overrightarrow{b}=x\hat{ı}+y\hat{ȷ}+z\hat{k}$

$\overrightarrow{a}\cdot \overrightarrow{b}=3\Rightarrow (\hat{ı}+\hat{ȷ}+\hat{k})\cdot (x\hat{ı}+y\hat{ȷ}+z\hat{k})=3$

$\Rightarrow x+y+z=3\cdots \left(1\right)$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{ȷ}& \hat{k}\\ 1& 1& 1\\ x& y& z\end{array}\right|=\hat{i}(z-y)-\hat{ȷ}(z-x)+\hat{k}(y-x)$

accarding to question

$\hat{1}(z-y)+\hat{ȷ}(x-z)+\hat{k}(y-x)=\hat{ȷ}-\hat{k}$

comparing both sides

$\begin{array}{c}z-y=0\Rightarrow z=y\cdots \left(2\right)\\ x-z=1\Rightarrow x=1+z\\ y-x=-1\Rightarrow y+1=x\cdots \left(3\right)\\ \text{using equation}\left(1\right),\left(2\right)\text{and}\left(3\right)\\ \cdot x+y+z=3\end{array}$

$\Rightarrow y+1+y+y=3\Rightarrow y=\frac{2}{3}$

$\therefore y=\frac{2}{3},z=\frac{2}{3},x=\frac{5}{3}$

$\overrightarrow{b}=\frac{5}{3}\hat{ı}+\frac{2}{3}\hat{ȷ}+\frac{2}{3}\hat{k}=\frac{1}{3}(5\hat{ı}+2\hat{ȷ}+2\hat{k})$

$\therefore$ Answer : option (A)