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If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\vec{j}-\vec{k}$, then $\vec{b}$ is
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The correct answer is:
$\hat{i}$
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\vec{j}-\vec{k}$,
Let $\bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \widehat{k}$
Now, $\hat{j}-\widehat{k}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 1 & 1 \\ b_2 & b_2 & b_3\end{array}\right|$
$\Rightarrow b_3-b_2=0, b_1-b_3=1, b_2-b_1=-1$
$\Rightarrow b_3=b_2, b_1=b_2+1$
Also, $\bar{a} \cdot \bar{b}=1$
$\Rightarrow b_1+b_2+b_3=1$
$\Rightarrow 3 b_2+1=1$
$\Rightarrow b_2=0$
$\Rightarrow b_1=1, b_3=0$
Thus, $\bar{b}=\hat{i}$
Let $\bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \widehat{k}$
Now, $\hat{j}-\widehat{k}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 1 & 1 \\ b_2 & b_2 & b_3\end{array}\right|$
$\Rightarrow b_3-b_2=0, b_1-b_3=1, b_2-b_1=-1$
$\Rightarrow b_3=b_2, b_1=b_2+1$
Also, $\bar{a} \cdot \bar{b}=1$
$\Rightarrow b_1+b_2+b_3=1$
$\Rightarrow 3 b_2+1=1$
$\Rightarrow b_2=0$
$\Rightarrow b_1=1, b_3=0$
Thus, $\bar{b}=\hat{i}$
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