• question_answer If b and c are any two non-collinear unit vectors and a is any vector, then $(\mathbf{a}\,.\,\mathbf{b})\,\mathbf{b}+(\mathbf{a}\,.\,\mathbf{c})\,\mathbf{c}+\frac{\mathbf{a}\,.\,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}\,(\mathbf{b}\times \mathbf{c})=$ [IIT 1996] A) a B) b C) c D) 0

Solution :

• Let $\mathbf{i}$ be a unit vector in the direction of $\mathbf{b},\,\mathbf{j}$ in the direction of $\mathbf{c}.$ Note that $\mathbf{b}=\mathbf{i}$ and $\mathbf{c}=\mathbf{j}$
• We have $\mathbf{b}\times \mathbf{c}=\,|\mathbf{b}||\mathbf{c}|\sin \alpha \,\mathbf{k}=\sin \alpha \,\mathbf{k}$, where $\mathbf{k}$ is a unit vector perpendicular to $\mathbf{b}$ and $\mathbf{c}.$
• $\Rightarrow \,|\mathbf{b}\times \mathbf{c}|\,=\sin \alpha \Rightarrow \mathbf{k}=\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}$          Any vector $\mathbf{a}$ can be written as a linear combination of $\mathbf{i},\,\,\mathbf{j}$ and $\mathbf{k}.$
• Let $\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}.$
• Now   $\mathbf{a}\,.\,\mathbf{b}=\mathbf{a}\,.\,\mathbf{i}={{a}_{1}},$ $\mathbf{a}\,.\,\mathbf{c}=\mathbf{a}\,.\,\mathbf{j}={{a}_{2}}$                                 and $\mathbf{a}\,.\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}=\mathbf{a}\,.\,\mathbf{k}={{a}_{3}}$
• Thus $(\mathbf{a}\,.\,\mathbf{b})\mathbf{b}+(\mathbf{a}\,.\,\mathbf{c})\mathbf{c}+\frac{\mathbf{a}\,.\,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}(\mathbf{b}\times \mathbf{c}|$
• $={{a}_{1}}\mathbf{b}+{{a}_{2}}\mathbf{c}+{{a}_{3}}\frac{(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}=\mathbf{a}$.

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