Question

If $a,b$ and $c$ are non-coplanar, then the value of $\frac{1}{3}\left[\frac{a·(b\times c)}{b·(c\times a)}\right]-\frac{1}{2}\left[\frac{b·(c\times a)}{c·(a\times b)}\right]$ is?

• A. $-\frac{1}{2}$
• B. $-\frac{1}{3}$
• C. $-\frac{1}{6}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C

$-\frac{1}{6}$

Find the value of $\frac{1}{3}\left[\frac{a·(b\times c)}{b·(c\times a)}\right]-\frac{1}{2}\left[\frac{b·(c\times a)}{c·(a\times b)}\right]$:

Given that, $a,b$ and $c$ are non-coplanar

$\frac{1}{3}\left[\frac{a·(b\times c)}{b·(c\times a)}\right]-\frac{1}{2}\left[\frac{b·(c\times a)}{c·(a\times b)}\right]$

$=\frac{a·(b\times c)}{3b·(c\times a)}-\frac{b·(c\times a)}{2c·(a\times b)}$

$=\frac{\left[abc\right]}{3\left[bca\right]}-\frac{\left[bca\right]}{2\left[cab\right]}$ $\left[\mathbf{\because }\mathit{a}\mathbf{·}\left(b\times c\right)\mathbf{}\mathbf{=}\mathbf{}\left[abc\right]\right]$

$=\frac{\left[abc\right]}{3\left[abc\right]}-\frac{\left[bca\right]}{2\left[bca\right]}$ $\left[\mathbf{\because }\left[bca\right]\mathbf{}\mathbf{=}\mathbf{}\left[abc\right]\mathbf{,}\mathbf{}\left[cab\right]\mathbf{}\mathbf{=}\mathbf{}\left[bca\right]\right]$

$=\frac{1}{3}-\frac{1}{2}$

$=\mathbf{-}\frac{\mathbf{1}}{\mathbf{6}}$

Hence, Option ‘C’ is Correct.