Let a→=j^−k^ and c→=i^−j^−k^ Then the vector b→satisfying a→×b→+c→=0, and a→⋅b→=3 is

a→×b→=−c→,a→×(a→×b→)=−a→×c→⇒ (a→⋅b→)a→−(a→⋅a→)b→=c→×a→∴ b→=1(a→⋅a→)[3a→+a→×c→]a→⋅a→=1+1=2.a→×c→=i^j^k^01−11−1−1=−2i^−j^−k^∴b→=12[3j^−3k^−2i^−j^−k^]=12(−2i^+2j^−4k^)=−i^+j^−2k^

Let a→=j^−k^ and c→=i^−j^−k^ Then the vector b→satisfying a→×b→+c→=0, and a→⋅b→=3 is

a→×b→=−c→,a→×(a→×b→)=−a→×c→⇒ (a→⋅b→)a→−(a→⋅a→)b→=c→×a→∴ b→=1(a→⋅a→)[3a→+a→×c→]a→⋅a→=1+1=2.a→×c→=i^j^k^01−11−1−1=−2i^−j^−k^∴b→=12[3j^−3k^−2i^−j^−k^]=12(−2i^+2j^−4k^)=−i^+j^−2k^

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Let $\vec{a} = \hat{j} - \hat{k} and \vec{c} = \hat{i} - \hat{j} - \hat{k}$ Then the vector $\vec{b}$
satisfying $\vec{a} \times \vec{b} + \vec{c} = 0, and \vec{a} \cdot \vec{b} = 3$ is

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$- \hat{i} + \hat{j} - 2 \hat{k}$

$2 \hat{i} - \hat{j} + 2 \hat{k}$

$\hat{i} - \hat{j} - 2 \hat{k}$

$\hat{i} + \hat{j} - 2 \hat{k}$

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Detailed Solution

$\vec{a} \times \vec{b} = - \vec{c}, \vec{a} \times (\vec{a} \times \vec{b}) = - \vec{a} \times \vec{c}$

$\begin{array}{l} \Rightarrow (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} = \vec{c} \times \vec{a} \\ ∴ \vec{b} = \frac{1}{(\vec{a} \cdot \vec{a})} [3 \vec{a} + \vec{a} \times \vec{c}] \\ \vec{a} \cdot \vec{a} = 1 + 1 = 2 . \end{array}$

$\begin{array}{l} \vec{a} \times \vec{c} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - 1 \\ 1 & - 1 & - 1 \end{array}| = - 2 \hat{i} - \hat{j} - \hat{k} \\ ∴ \vec{b} = \frac{1}{2} [3 \hat{j} - 3 \hat{k} - 2 \hat{i} - \hat{j} - \hat{k}] \\ = \frac{1}{2} (- 2 \hat{i} + 2 \hat{j} - 4 \hat{k}) = - \hat{i} + \hat{j} - 2 \hat{k} \end{array}$