Question

If $\alpha =p(b\times c)+q(c\times a)+r(a\times b)$ and $\left[abc\right]=\frac{1}{p+q+r}$, then find the value of $\alpha .(a+b+c).$

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

1

$\alpha .a=p\left[abc\right]+0+0=p\left[abc\right]$
$\alpha .b=q\left[abc\right]+0+0=q\left[abc\right]$
$\alpha .c=r\left[abc\right]+0+0=r\left[abc\right]$
$\therefore \alpha \cdot (a+b+c)=\sum \alpha .a=(p+q+r)\left[abc\right]=1$ (by given relation)