Question

If $r=\alpha b\times c+\beta c\times a+\gamma a\times b$ and $\left[abc\right]=2$, then $\alpha +\beta +\gamma$ is equal to

• A. $r·[b\times c+c\times a+a\times b]$
• B. $\frac{1}{2}r·(a+b+c)$
• C. $2r·(a+b+c)$
• D. $4$

Answer 1

The correct option is B

$\frac{1}{2}r·(a+b+c)$

Explanation for the correct option.

Step 1: Prerequisites for the solution

Here, we need to determine $\left(r·\alpha +r·\beta +r·\gamma \right)$

$r.a=\alpha \left(a.b\times c\right)+\beta \left(a.c\times a\right)+\gamma \left(a.a\times b\right)$

As per the vector multiplication $\left(a·c\times a\right)\text{and}\left(a·a\times b\right)$ will be zero then

$\begin{array}{rcl}r.a& =& \alpha \left(a.b\times c\right)+0+0\\ & =& \alpha \left[abc\right]\end{array}$

Similarly,

$\begin{array}{rcl}r.b& =& \beta \left[abc\right]\text{and}r.c=\gamma \left[abc\right]\end{array}$

From this, we have

$r·a+r·b+r·c=\alpha \left[abc\right]+\beta \left[abc\right]+\gamma \left[abc\right]$

Step 2: Calculation for the required expression

Since $\left[abc\right]=2$,

$\begin{array}{rcl}\therefore r·\left(a+b+c\right)& =& 2\alpha +2\beta +2\gamma \\ & \Rightarrow & r·\left(a+b+c\right)=2\left(\alpha +\beta +\gamma \right)\\ & \Rightarrow & \alpha +\beta +\gamma =\frac{1}{2}r·\left(a+b+c\right)\end{array}$

Hence, option B is correct.