A) \[8\mathbf{d}\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
B) \[8\mathbf{d}\,\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
C) \[\frac{\mathbf{d}\,}{8}.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
D) \[\frac{\mathbf{d}\,}{8}\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
Correct Answer: A
Solution :
\[\mathbf{d}\,.\,\mathbf{c}=\lambda (\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c}+\mu (\mathbf{b}\times \mathbf{c})\,.\,\mathbf{c}+\nu (\mathbf{c}\times \mathbf{a})\,.\mathbf{c}\] \[=\lambda \,[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]+0+0=\lambda \,[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{\lambda }{8}\] Hence \[\lambda =8(\mathbf{d}\,.\,\mathbf{c}),\] \[\mu =8(\mathbf{d}\,.\,\mathbf{a})\] and \[\nu =8(\mathbf{d}\,.\,\mathbf{b})\] Therefore, \[\lambda +\mu +\nu =8\mathbf{d}\,.\,\mathbf{c}+8\mathbf{d}.\,\mathbf{a}+8\mathbf{d}\,.\,\mathbf{b}\] \[=8\mathbf{d}\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c}).\]
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