Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Kinda late and my brain is glitching.

I know this is wrong, but I can't give a good reason why. I gave a counterexample to prove it is false, but I don't understand why this is wrong.

$$\frac{u \cdot v}{|u|} = \frac{v \cdot u}{|v|}$$

$$\hat{u} \cdot v \neq v \cdot \hat{u}$$

$u$ and $v$ are nonzero.

share|improve this question
    
Does $\hat{u}$ mean normalized version of $u$? –  sdcvvc May 12 '12 at 5:47
    
$u$ and $v$ are vectors and $.$ is dot product? –  Dilawar May 12 '12 at 5:47
    
Yes, to both question. –  Hawk May 12 '12 at 5:50

3 Answers 3

up vote 3 down vote accepted

$\hat{u}\cdot v \neq v \cdot \hat{u}$ doesn't make sense. The dot product is commutative. Since $v$ and $\hat{u}$ are two vectors, it should be such that $\hat{u}\cdot v = v\cdot \hat{u}$.

The first statement cannot be true unless $u=v$. ($u=v$ would still not imply $\hat{u}\cdot v \neq v \cdot \hat{u}$.)

share|improve this answer

This does not make sense because though $u\cdot v=v\cdot u$, the denominators are different. For the second line how do you distinguish inner product and scalar product?

share|improve this answer

In one dimension, vectors are reals, and $\frac{u}{|u|}=\operatorname{sgn} u$ (where sgn is 1 for positive, -1 for negative).

It should be intuitively obvious that

$(\operatorname{sgn} x) y$ and $x (\operatorname{sgn} y)$

are different things; the length of the first is $|y$| while length of the second is $|x|$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.