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If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{a} \cdot \mathbf{b}=1$ and $\mathbf{a} \times \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$, then $\mathbf{b}=$
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Verified Answer
The correct answer is:
$\hat{\mathbf{i}}$
Let vector $\mathbf{b}=b_1 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+b_3 \hat{\mathbf{k}}$, then
$\mathbf{a} \cdot \mathbf{b}=b_1+b_2+b_3=1$ (given )
and $\begin{aligned} \mathbf{a} \times \mathbf{b}=\left(b_3-b_2\right) \hat{\mathbf{i}}+\left(b_1-b_3\right) \hat{\mathbf{j}}+ & \left(b_2-b_1\right) \hat{\mathbf{k}} \\ & =\hat{\mathbf{j}}-\hat{\mathbf{k}} \text { (given) }\end{aligned}$
So,
$b_2=b_3, b_1=b_3+1$ and $b_1=b_2+1$
From Eqs. (i) and (ii), we get
$\left(b_3+1\right)+b_3+b_3=1$
$\Rightarrow \quad b_3=0=b_2$ and $b_1=1$
So,
$\mathbf{b}=\hat{\mathbf{i}}$
Hence, option (d) is correct.
$\mathbf{a} \cdot \mathbf{b}=b_1+b_2+b_3=1$ (given )
and $\begin{aligned} \mathbf{a} \times \mathbf{b}=\left(b_3-b_2\right) \hat{\mathbf{i}}+\left(b_1-b_3\right) \hat{\mathbf{j}}+ & \left(b_2-b_1\right) \hat{\mathbf{k}} \\ & =\hat{\mathbf{j}}-\hat{\mathbf{k}} \text { (given) }\end{aligned}$
So,
$b_2=b_3, b_1=b_3+1$ and $b_1=b_2+1$
From Eqs. (i) and (ii), we get
$\left(b_3+1\right)+b_3+b_3=1$
$\Rightarrow \quad b_3=0=b_2$ and $b_1=1$
So,
$\mathbf{b}=\hat{\mathbf{i}}$
Hence, option (d) is correct.
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