Question

If $\alpha =x(a\times b)+y(b\times c)+z(c\times a)$ and $[abc]=\frac{1}{8}$ then $x+y+z$ is equal to?

• A. $8\alpha ·(a+b+c)$
• B. $\alpha ·(a+b+c)$
• C. $8(a+b+c)$
• D. None of these

Answer 1

The correct option is A

$8\alpha ·(a+b+c)$

Explanation for the correct option:

Step 1. Find the value of $x+y+z$:

Given, $\alpha =x(a\times b)+y(b\times c)+z(c\times a)$

$\Rightarrow$ $\alpha ·a=x(a\times b)·a+y(b\times c)·a+z(c\times a)·a$

$\Rightarrow$ $\alpha ·a=0+y(b\times c)·a+0$

$\Rightarrow$ $\alpha ·a=y\left[abc\right]$

Step 2. Find the values of $x,y,z$:

$\because \left[abc\right]=\frac{1}{8}$

$\begin{array}{rcl}\therefore y& =& \frac{\alpha ·a}{\left[abc\right]}\\ & =& 8\alpha ·a\end{array}$

Similarly,

$\begin{array}{rcl}\alpha ·b& =& z(c\times a)·b\\ & =& z\left[abc\right]\end{array}$

$\Rightarrow$$z=8\alpha ·b$

Again,

$\begin{array}{rcl}\alpha ·c& =& x(a\times b)·c\\ & =& x\left[abc\right]\end{array}$

$\Rightarrow$$x=8\alpha ·c$

Step 3. Put the values of $x,y,z$ in given expression, we get

$\begin{array}{rcl}\therefore x+y+z& =& 8\alpha ·c+8\alpha ·a+8\alpha ·b\\ & =& \mathbf{}\mathbf{8}\mathit{\alpha }\mathbf{}\mathbf{·}\mathbf{}\mathbf{(}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathbf{)}\end{array}$

Hence, Option ‘A’ is Correct.