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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.

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

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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?

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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 $|x|$.

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