Question

If vector$(a+2b+3c)=0,$ then vector $(a\times b+b\times c+c\times a)$ is equal to

• A. $6(b\times c)$
• B. $3(b\times c)$
• C. $2(b\times c)$
• D. $0$

Answer 1

The correct option is A

$6(b\times c)$

Explanation for the correct option:

Finding the value of the expression:

Given,

$(a+2b+3c)=0$

Multiply by $\overrightarrow{b}$.

$$\begin{gathered} \overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0 \\ \overrightarrow{a}x\overrightarrow{b}+2\overrightarrow{b}x\overrightarrow{b}+3\overrightarrow{c}x\overrightarrow{b}=0 \\ \Rightarrow \overrightarrow{a}x\overrightarrow{b}+3\overrightarrow{c}x\overrightarrow{b}=0\left[\because b\times b=0\right] \\ \Rightarrow \overrightarrow{a}x\overrightarrow{b}=3\overrightarrow{b}x\overrightarrow{c}...\left(1\right) \end{gathered}$$

In a similar manner, multiply by $\overrightarrow{a}$ in the given equation.

$$\begin{gathered} \overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0 \\ \Rightarrow \overrightarrow{a}x\overrightarrow{a}+2\overrightarrow{b}x\overrightarrow{a}+3\overrightarrow{c}x\overrightarrow{a}=0 \\ \Rightarrow \overrightarrow{c}\times \overrightarrow{a}=\frac{2}{3}\overrightarrow{a}\times \overrightarrow{b}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{a}\mathbf{\times }\mathbf{b}\mathbf{=}\mathbf{-}\mathbf{b}\mathbf{\times }\mathbf{a}\mathbf{]} \\ \Rightarrow =\frac{2}{3}\times 3\left(\overrightarrow{b}x\overrightarrow{c}\right)\left[\mathrm{from}1\right] \\ \Rightarrow =2\left(\overrightarrow{b}x\overrightarrow{c}\right)...\left(2\right) \end{gathered}$$

From $\left(1\right)$ and $\left(2\right)$6

$\overrightarrow{a}x\overrightarrow{b}+\overrightarrow{b}x\overrightarrow{c}+\overrightarrow{c}x\overrightarrow{a}=6\left(\overrightarrow{b}x\overrightarrow{c}\right)$

Hence, the correct option is A.