# Simple dot product question

u and v are vectors

\begin{align*} \|u+v\| &= 4 \\ \|u\| &= \|v\| = 3\\ \end{align*}

find $u.v$

$u.u= \|u\|^2 \ = 9 and\ v.v=\|v\|^2 \ and (u+v).(u+v)=u.u+2u.v+v.v$

so $u.v = \frac{(16-9-9)}{ 2} = -1$

My question is why is $u.v$ not equal to $9$? Im sorry I have forgotten the basics.

• $u \cdot u = \|u\|^2 = 9$ Commented Apr 15, 2013 at 7:27
• @user1730308 you have forgotten to take the square root:$\|u\|=\sqrt{u.u}$ Commented Apr 15, 2013 at 7:29
• sorry im starting to get sloppy, just fixed it Commented Apr 15, 2013 at 7:32
• sorry i meant why is u.v not equal to 9 Commented Apr 15, 2013 at 7:33

Well there's no "intuitive" reason as to why, because at least as far as I know it dot product is just a specific case of an inner product, one can have many different inner products products defined as long as they follow some rules:

Over $\mathbb{R}$ :

An inner product $(,):V\times V\to\mathbb{R}$:

1. Linear with respect to the first parameter: $\left(u+u^{'},v\right) = \left(u,v\right) +\left(u^{'},v\right)$.
2. Homogeneous with respect to the first parameter: goven $a \in \Bbb{R}$ then $\left(au,v\right) = a\left(u,v\right)$.
3. Commutative : $\left(u,v\right) = \left(v,u\right)$
4. Positive $u\in V , u\neq 0_{V} \Rightarrow \left(u,u\right) >0$

A bit different Over $\mathbb{C}$ :

All first three are the same then there's a slight difference:

• Conjugate symmetry : $\left(u,v\right) = \overline{\left(v,u\right)}$
• Positive : $u\in V , u\neq 0_{V} \Rightarrow \left(u,u\right) >0$ which actually contains two properties :
• The dot product of a non zero vector with itself is real.
• The dot product of a non zero vector with itself is positive

So you actually can define a lot of different inner product products and what you're specifically talking about is the standard inner product (at least that the way my linear algebra prof calls it :) ) which we commonly use.

As to a specific example: what you did what correct, but consider the vectors :$v = \left(\begin{matrix}\frac{3}{\sqrt{2}}\\ \frac{3}{\sqrt{2}} \end{matrix}\right) , u = \left(\begin{matrix}- \frac{3}{\sqrt{2}}\\ \frac{3}{\sqrt{2}} \end{matrix}\right)$ in $\mathbb{R}^2$

You can see that $||u|| = ||v|| = 3$ but $u\cdot v = 0$ so the most intuitive thing I can propose is to think about the fact that there are many vectors with the same length but that does not imply anything specific on their coordinates , so their dot product can be different then their individual inner product.

• very helpful! i appreciate it Commented Apr 15, 2013 at 9:06

$\|u\|=\sqrt{u.u}$. Now, $$(u+v)\cdot (u+v)=u\cdot v+2u\cdot v+v\cdot v=\|u\|^2+2u\cdot v+\|v\|^2=18+2u\cdot v$$ It follows $$u\cdot v=\frac{1}{2}[(u+v)\cdot (u+v)-18]=\frac{1}{2}[\|u+v\|^2-18]=\frac{1}{2}(16-18)=-1$$

Norm is not multiplicative: $\|u\cdot v\|\neq \|u\|\cdot\|v\|$, it is only homogeneous: $\|\lambda v\|=|\lambda|\|v\|$, for every scalar $\lambda$

• Ok one last question how did the u vector become u+v? Commented Apr 15, 2013 at 8:03
• @user1730308 it doesen't become $u+v$. First line of my answer is only the definition of norm in terms of scalar product , as an example i wrote the norm of $u$, but you can put any vector in place of $u$. Then line 2 uses the bilinearity of scalar product. Commented Apr 15, 2013 at 8:07
• got it thanks for the help! Commented Apr 15, 2013 at 9:06