Question

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

• A. $d\cdot (a+b+c)$
• B. $2d\cdot (a+b+c)$
• C. $4d\cdot (a+b+c)$
• D. $8d\cdot (a+b+c)$

Answer 1

The correct option is D $8d\cdot (a+b+c)$

$\overline{d}.\overline{a}=\lambda \left(0\right)+\frac{4}{8}+\left(0\right)\nu ,\left[abc\right]=\frac{1}{8}$
$\overline{d}.\overline{b}=\lambda \left(0\right)+4\left(0\right)+\left[abc\right]\nu$
$\overline{d}.\overline{c}=\lambda (\overline{a}\times \overline{b}).\overline{c}$
$=\lambda \left(\overline{a}\overline{b}\overline{c}\right)$

$=\frac{\lambda }{8}$
$\Rightarrow \lambda +4+\nu =8\overline{d}.(\overline{a}+\overline{b}+\overline{c})$