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Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. Then vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $\vec{a} \cdot \vec{b}=3$ is
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The correct answer is:
$-\hat{i}+\hat{j}-2 \hat{k}$
$-\hat{i}+\hat{j}-2 \hat{k}$
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}$
$\Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0$
$\Rightarrow\left(\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{\mathrm{k}}\right) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})=0$
$\mathrm{~b}_1-\mathrm{b}_2-\mathrm{b}_3=0$
and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3$
$\Rightarrow \mathrm{b}_2-\mathrm{b}_3=3$
$\mathrm{~b}_1=\mathrm{b}_2+\mathrm{b}_3=3+2 \mathrm{~b}_3$
$\overrightarrow{\mathrm{b}}=\left(3+2 \mathrm{~b}_3\right) \hat{\mathrm{i}}+\left(3+\mathrm{b}_3\right) \hat{\mathrm{j}}+\mathrm{b}_3 \hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0$
$\Rightarrow\left(\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{\mathrm{k}}\right) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})=0$
$\mathrm{~b}_1-\mathrm{b}_2-\mathrm{b}_3=0$
and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3$
$\Rightarrow \mathrm{b}_2-\mathrm{b}_3=3$
$\mathrm{~b}_1=\mathrm{b}_2+\mathrm{b}_3=3+2 \mathrm{~b}_3$
$\overrightarrow{\mathrm{b}}=\left(3+2 \mathrm{~b}_3\right) \hat{\mathrm{i}}+\left(3+\mathrm{b}_3\right) \hat{\mathrm{j}}+\mathrm{b}_3 \hat{\mathrm{k}}$
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