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

$2$

Explanation for the correct option:

Step1. Find the value of $\mathit{a}\mathbf{\times }\mathit{a}$:

Given,$a+2b+3c=0$

We can write, $a=-(2b+3c)$

$a\times a=-\left(2\right(b\times a)+3(c\times a\left)\right)$

Since,$axa=0$

$\Rightarrow (2b\times a+3(c\times a\left)\right)=0\dots \dots \dots .\left(i\right)$

Step2. Find the value of $\mathit{b}\mathbf{\times }\mathit{b}$:

Again, we can write, the given expression as,

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

Adding equation (i) and (ii), we get;

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

​Step 3. Apply the rule $\mathit{a}\mathbf{\times }\mathit{b}\mathbf{=}\mathbf{-}\mathbf{(}\mathit{b}\mathbf{\times }\mathit{a}\mathbf{)}$:

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

Now, as per the given question,

$$\begin{gathered} \Rightarrow a\times b+b\times c+c\times a=ka\times b \\ \Rightarrow 1=2a\times b=ka\times b\mathbf{}\mathbf{(}\mathbf{}\mathbf{f}\mathbf{r}\mathbf{o}\mathbf{m}\mathbf{}\mathbf{e}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{\left(}\mathbf{i}\mathbf{i}\mathbf{i}\mathbf{\right)}\mathbf{)} \end{gathered}$$

Thus, $k=2$

Hence the correct option is C.