A) \[\mathbf{a},\,\,\mathbf{a}-\mathbf{x}\]
B) \[\mathbf{a}-\mathbf{b},\,\,\mathbf{b}\]
C) \[\mathbf{b},\,\,\mathbf{a}-b\]
D) None of these
Correct Answer: D
Solution :
Multiplying (i) scalarly by \[\mathbf{a},\] we get \[\mathbf{a}\,.\,\mathbf{x}+\mathbf{a}\,.\,\mathbf{y}={{\mathbf{a}}^{2}}\] \[\therefore \,\mathbf{a}\,.\,\mathbf{y}={{\mathbf{a}}^{2}}-1\] ?..(iv), {By (iii)} Again \[\mathbf{a}\times (\mathbf{x}\times \mathbf{y})=\mathbf{a}\times \mathbf{b}\]or \[(\mathbf{a}\,.\,\mathbf{y})\mathbf{x}-(\mathbf{a}\,.\,\mathbf{x})\mathbf{y}=\mathbf{a}\times \mathbf{b}\] \[({{\mathbf{a}}^{2}}-1)\mathbf{x}-\mathbf{y}=\mathbf{a}\times \mathbf{b}\] ?..(v), {By (iii) and (iv)} Adding and subtracting (i) and (v), we get \[x=\frac{\mathbf{a}+(\mathbf{a}\times \mathbf{b})}{{{\mathbf{a}}^{2}}}\] and \[\mathbf{y}=\mathbf{a}-\mathbf{x}\] etc.
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