Question

If $[a\times bb\times cc\times a]=\lambda {\left[abc\right]}^2$, then $\lambda$ is equal to?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Find the value of $\lambda$:

Given, $[a\times bb\times cc\times a]=\lambda {\left[abc\right]}^2$

Take L.H.S of the equation

$[a\times bb\times cc\times a]=(a\times b)·\left[(b\times c)\times (c\times a)\right]$

$=(a\times b)\left(\right[bca]c-[bcc\left]a\right)$ $\left[\mathbf{\because }\mathbf{(}\mathit{a}\mathbf{\times }\mathit{b}\mathbf{)}\mathbf{\times }\mathbf{(}\mathit{c}\mathbf{\times }\mathit{d}\mathbf{)}\mathbf{=}\mathbf{\left[}\mathit{a}\mathit{b}\mathit{d}\mathbf{\right]}\mathit{c}\mathbf{-}\mathbf{\left[}\mathit{a}\mathit{b}\mathit{c}\mathbf{\right]}\mathit{d}\right]$

$=(a\times b)\left(\right[bca\left]c\right)$ $\left[\mathbf{\because }\mathbf{\left[}\mathit{b}\mathit{c}\mathit{c}\mathbf{\right]}\mathbf{=}\mathbf{0}\right]$

$=(a\times b)·c\left[bca\right]$

$=\left[abc\right]\left[bca\right]$

$={\left[abc\right]}^2$ $\left[\mathbf{\because }\mathbf{\left[}\mathit{b}\mathit{c}\mathit{a}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathit{a}\mathit{b}\mathit{c}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathit{c}\mathit{a}\mathit{b}\mathbf{\right]}\right]$

$=$R.H.S

$\mathbf{\therefore }\mathit{\lambda }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘B’ is Correct.