Question

If $d=\lambda (a\times b)+\mu (b\times c)+v(c\times a),\left[abc\right]=\frac{1}{8}$ and $d.(a+b+c)=8$ then, $\lambda +\mu +v$ is equal to (where $\left[abc\right]$denotes scalar triple product.)

Answer 1

$d\cdot c=\lambda (a\times b)\cdot c+\mu (b\times c)\cdot c+v(c\times a)\cdot c$
$=\lambda \left[abc\right]+0+0=\lambda \left[abc\right]=\frac{\lambda }{8}$.
Hence $\lambda =8(d\cdot c).$
Similarly, $\mu .=8(d\cdot a)$ and
$v=8(d\cdot b)$
$\therefore \lambda +\mu +v=8d\cdot c+8d\cdot a+8d\cdot b$
$=8d\cdot (a+b+c)=64.$