Question

Prove that for any three vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ , $\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{b}+\overrightarrow{c} \right)\left( \overrightarrow{c}+\overrightarrow{a} \right) \right]=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]$ .

Answer 1

Hint: Consider the Left Hand Side of the equality and use the result $\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}=\overrightarrow{a}.\left[ \overrightarrow{b}\times \overrightarrow{c} \right]$.
To expand it and then further simplify it by expanding the cross product . After that try to make the maximum of the terms equal to 0 as the right hand side of the equality involves only one term.
And hence with little manipulations we can get the desired equality.

Complete step by step answer:
We will first consider the Left Hand Side of the equality , i.e.
$\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{b}+\overrightarrow{c} \right)\left( \overrightarrow{c}+\overrightarrow{a} \right) \right]$
We know from a basic result that
$\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}=\overrightarrow{a}.\left[ \overrightarrow{b}\times \overrightarrow{c} \right]$
Applying this result in the above Left Hand Side of the equality, we get,
$\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{b}+\overrightarrow{c} \right)\left( \overrightarrow{c}+\overrightarrow{a} \right) \right]=\left( \overrightarrow{a}+\overrightarrow{b} \right).\left[ \left( \overrightarrow{b}+\overrightarrow{c} \right)\times \left( \overrightarrow{c}+\overrightarrow{a} \right) \right]$
Now, expanding the cross product in the last step, we get,
$=\left( \overrightarrow{a}+\overrightarrow{b} \right).\left[ \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\left( \overrightarrow{b}\times \overrightarrow{a} \right)+\left( \overrightarrow{c}\times \overrightarrow{c} \right)+\left( \overrightarrow{c}\times \overrightarrow{a} \right) \right]$
Now, since $\overrightarrow{c}\times \overrightarrow{c}=\left| \overrightarrow{c} \right|\left| \overrightarrow{c} \right|\operatorname{Sin}0\widehat{n}=0$
Put, this value to the last step, we get
$\begin{align}
  & =\left( \overrightarrow{a}+\overrightarrow{b} \right).\left[ \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\left( \overrightarrow{b}\times \overrightarrow{a} \right)+0+\left( \overrightarrow{c}\times \overrightarrow{a} \right) \right] \\
 & =\left( \overrightarrow{a}+\overrightarrow{b} \right).\left[ \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\left( \overrightarrow{b}\times \overrightarrow{a} \right)+\left( \overrightarrow{c}\times \overrightarrow{a} \right) \right] \\
\end{align}$
Now, expanding the dot product in the last step, we get,
$=\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)+\overrightarrow{b}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)+\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{a} \right)+\overrightarrow{b}.\left( \overrightarrow{b}\times \overrightarrow{a} \right)+\overrightarrow{a}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)+\overrightarrow{b}.\left( \overrightarrow{c}\times \overrightarrow{a} \right)$
Now, again using the previous result i.e.
 $\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}=\overrightarrow{a}.\left[ \overrightarrow{b}\times \overrightarrow{c} \right]$
Applying this to the last step, we get
$=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+\left[ \overrightarrow{b}\overrightarrow{b}\overrightarrow{c} \right]+\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{a} \right]+\left[ \overrightarrow{a}\overrightarrow{c}\overrightarrow{a} \right]+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right]$
Since, $\left[ \overrightarrow{b}\overrightarrow{b}\overrightarrow{c} \right]=\left[ \overrightarrow{c}\overrightarrow{b}\overrightarrow{b} \right]=\overrightarrow{c}.\left( \overrightarrow{b}\times \overrightarrow{b} \right)$
And we know that $\left( \overrightarrow{b}\times \overrightarrow{b} \right)=0$
$\begin{align}
  & \Rightarrow \left[ \overrightarrow{b}\overrightarrow{b}\overrightarrow{c} \right]=\left[ \overrightarrow{c}\overrightarrow{b}\overrightarrow{b} \right] \\
 & =\overrightarrow{c}.\left( \overrightarrow{b}\times \overrightarrow{b} \right) \\
 & =\overrightarrow{c}.0 \\
 & =0
\end{align}$
Similarly, we can say that
$\begin{align}
  & \left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{a} \right]=0 \\
 & \left[ \overrightarrow{b}\overrightarrow{b}\overrightarrow{a} \right]=0 \\
 & \left[ \overrightarrow{a}\overrightarrow{c}\overrightarrow{a} \right]=0 \\
\end{align}$
Now, using these results, the Left Hand Side now becomes,
$\begin{align}
  & =\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+0+0+0+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right] \\
 & =\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+\left[ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right] \\
 & =\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]+\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right] \\
 & =2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right] \\
\end{align}$
=Right Hand Side

Hence,
$\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{b}+\overrightarrow{c} \right)\left( \overrightarrow{c}+\overrightarrow{a} \right) \right]=2\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]$

Note: The possibility of mistake here is that by considering the Left Hand Side of the equality some student may expand it like the ordinary multiplication or vectors that means some could have used the distributive property of usual multiplication for expanding $\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\left( \overrightarrow{b}+\overrightarrow{c} \right)\left( \overrightarrow{c}+\overrightarrow{a} \right) \right]$
But it is not an ordinary multiplication, it is a dot product of three vectors. So, to expand it
we have to use the $\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}=\overrightarrow{a}.\left[ \overrightarrow{b}\times \overrightarrow{c} \right]$ result and proceed the same.