If $\mathbf{d}=\lambda \,(\mathbf{a} imes \mathbf{b})+\mu \,(\mathbf{b} imes \mathbf{c})+
u \,(\mathbf{c} imes \mathbf{a})$and $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},$ then $\lambda +\mu +
u $ is equal to

Question

If $\mathbf{d}=\lambda \,(\mathbf{a}	imes \mathbf{b})+\mu \,(\mathbf{b}	imes \mathbf{c})+
u \,(\mathbf{c}	imes \mathbf{a})$and $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},$ then $\lambda +\mu +
u $ is equal to

Accepted Answer

$8\mathbf{d}\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})$

Answer

$8\mathbf{d}\,	imes \,(\mathbf{a}+\mathbf{b}+\mathbf{c})$

Answer

$\frac{\mathbf{d}\,}{8}.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})$

Answer

$\frac{\mathbf{d}\,}{8}	imes \,(\mathbf{a}+\mathbf{b}+\mathbf{c})$

vectors-m

Question: If $\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf{b}\times \mathbf{c})+\nu \,(\mathbf{c}\times \mathbf{a})$and $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},$ then $\lambda +\mu +\nu $ is equal to

vectors-m

Question: If $\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf{b}\times \mathbf{c})+\nu \,(\mathbf{c}\times \mathbf{a})$and $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},$ then $\lambda +\mu +\nu $ is equal to

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