My attempt:


Expand $ \langle (\textbf{v}+\textbf{w}), (\textbf{v}-\textbf{w}) \rangle = \textbf{v} \cdot (\textbf{v}-\textbf{w}) + \textbf{w} \cdot (\textbf{v}-\textbf{w}) = \textbf{v} \cdot \textbf{v} - \textbf{v} \cdot \textbf{w} + \textbf{w} \cdot \textbf{v} - \textbf{w} \cdot \textbf{w} = 1 - 1 + 1 - 1 = 0$

Which mean v+w and v-w are orthogonal.

(2) Are $\textbf{v} + \textbf{w}$ and $\textbf{v} - \textbf{w}$ unit vectors?

Not necessarily,

Find a counter example:


$\textbf{v} = \textbf{w} = \begin{pmatrix} 1\\0 \end{pmatrix} $


$ \langle (\textbf{v}+\textbf{w}), (\textbf{v}+\textbf{w}) \rangle = (v_1 + w_1)(v_1+w_1) + (v_2+w_2)(v_2+w_2) = 4$

So $\textbf{v}+\textbf{w}$ is not a unit vector.

Is (1) and (2) good enough proofs? Lastly how do I determine if v-w is a unit vector?

  • $\begingroup$ That's OK so far. Experiment with some unit vectors to get an idea on what v-w might be. $\endgroup$ Oct 31, 2016 at 9:21
  • $\begingroup$ In the first line you use that $v \cdot w = 1$, this does not seem to hold in gerenal. Consider $v=(1,0)$ and $w=(0,1)$. $\endgroup$
    – flawr
    Oct 31, 2016 at 9:21
  • $\begingroup$ Your proof looks correct. What I suggest is not necessary at all. But still I suggest to do the following: take a pen and paper, draw two unit vectors, draw the sum and difference of these vectors. $\endgroup$
    – lesnik
    Oct 31, 2016 at 9:38

2 Answers 2


It looks fine to me and enough. As for the last question:

$$\langle v-w,\,v-w\rangle=\left\|v\right\|^2+\left\|w\right\|^2-2\,\text{Im}\langle v,w\rangle\,i=2\left(1-\text{Im}\langle v,w\rangle\,i\right)$$

Now I think you can find a counter example for the above being a unitary vector (or perhaps you already did...?)


$\textbf{w} \cdot \textbf{v} \ne 1$ and $\textbf{v} \cdot \textbf{w} \ne 1$ necessarily as the angle between the 2 unit vectors is not specified. Rather, it is the case that $\textbf{w} \cdot \textbf{v} = \textbf{v} \cdot \textbf{w}$. Hence the two terms cancel off in the (1) question and the resultant dot product comes as zero.

Also when you are asked whether $\textbf{v} + \textbf{w}$ is a unit vector or not, after computing the self-dot product you should also mention that the magnitude of $\textbf{v} + \textbf{w}$ is $\sqrt(4) = 2$ (if you want to be rigorous in your proof).

And similarly for providing a counter-example for showing that $\textbf{v} - \textbf{w}$ is not a unit vector, you can take the 2 vectors $\begin{pmatrix} 1\\0 \end{pmatrix} $ and $\begin{pmatrix} -1\\0 \end{pmatrix} $ as $\textbf{v}$ and $\textbf{w}$ respectively.


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