Question

If $a+2b+3c=0$, then $a\times b+b\times c+c\times a=ka\times b,$
Where $k$ is equal to ?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given,

$\begin{array}{c}a+2b+3c=0\\ a=-\left(2b+3c\right)\\ a\times a=-\left(2\left(b\times a\right)+3\left(c\times a\right)\right)\\ \Rightarrow 0=-\left(2b\times a+3\left(c\times a\right)\right)..........\left(i\right)\end{array}$

Again,

$\begin{array}{c}2b=-\left(a+3c\right)\\ 2b\times b=-\left(a\times b+3c\times b\right)\\ \Rightarrow 0=-\left(a\times b+3c\times b\right)..........\left(ii\right)\end{array}$

equation (i) and (ii)

$\begin{array}{c}-\left(2b\times a+3c\times a\right)+\left(a\times b\right)+3c\times b=0\\ \Rightarrow 3a\times b-3c\times a-3b\times c=0\end{array}$

We know that,

$a\times b=-b\times a$

$a\times b=c\times a+b\times c...............\left(iii\right)$

Now,$ATQ;$

$a\times b+b\times c+c\times a=Ka\times b$

$1=2a\times b=ka\times b$ {from equation (iii)}

$\therefore k=2$

Option $C$ is correct answer.