This question already has an answer here:

The Homework Exercise I am working on, is:

Let $\overrightarrow{a}, \overrightarrow{b}$ be vectors. Show that $\overrightarrow{a} \cdot \overrightarrow{b}=\frac{1}{4}\left(\Vert{\overrightarrow{a}+\overrightarrow{b}\Vert^2}-\Vert{\overrightarrow{a}-\overrightarrow{b}\Vert^2}\right)$.

Things I tried:

  1. Using the property $\Vert \overrightarrow{u}\Vert^2=\overrightarrow{u} \cdot \overrightarrow{u}$, I tried to make $\Vert{\overrightarrow{a}+\overrightarrow{b}\Vert^2} =\left(\overrightarrow{a}+\overrightarrow{b} \right) \cdot \left(\overrightarrow{a}+\overrightarrow{b} \right)$ and $\Vert{\overrightarrow{a}-\overrightarrow{b}\Vert^2} =\left(\overrightarrow{a}-\overrightarrow{b} \right) \cdot \left(\overrightarrow{a}-\overrightarrow{b} \right)$

I'm not sure if that was a correct step, but then I substituted into the main equation to get $$ \overrightarrow{a} \cdot \overrightarrow{b} = \frac{1}{4}\left( (\overrightarrow{a}+\overrightarrow{b} ) \cdot (\overrightarrow{a}+\overrightarrow{b} ) - (\overrightarrow{a}-\overrightarrow{b} ) \cdot (\overrightarrow{a}-\overrightarrow{b} ) \right) $$

Am not sure where to go from here, or if I'm even heading in the right direction...


marked as duplicate by Jack, MrYouMath, Namaste linear-algebra Oct 17 '17 at 16:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Next use the distributive property of the dot product over vector addition, and collect like terms. $\endgroup$ – hardmath Oct 17 '17 at 12:29
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Polarization_identity $\endgroup$ – Jack Oct 17 '17 at 12:49
  • $\begingroup$ The proposed duplicate focuses on discovery of the polarization identity (or identities, if complex normed vector spaces are included), so it doesn't make an ideal duplicate. There are previous Questions with much the same focus as this one, but the Answers given there are in most cases less detailed than the calculation already provided in the problem statement here. So I'm voting to leave open. $\endgroup$ – hardmath Oct 17 '17 at 15:55

Use the fact that $$\vec a\cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a\cdot \vec c$$

and write it all out. A lot of things should cancel out.


Just expand the dot products by didtributivity in the right-hand side: \begin{align} (\vec a+\vec b)\cdot(\vec a+\vec b)-(\vec a-\vec b)\cdot(\vec a-\vec b)&=\begin{aligned}[t]\vec a\cdot\vec a&+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b\\ &-\vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a-\vec b\cdot\vec b \end{aligned}\\ &=2\vec a\cdot\vec b+2\vec b\cdot\vec a=4\vec a\cdot\vec b \end{align}


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