# The Transitive Property. Same relation or not?

Consider $\mathbf{u}, \mathbf{v}, \mathbf{w}$ vectors.

If $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, that is $\mathbf{u} \cdot \mathbf{v} = 0$, and the vectors $\mathbf{v}$ and $\mathbf{w}$ are parallel, that is $\mathbf{v} = c\mathbf{w}$ for some $c$, then by transitivity we can conclude that

$\mathbf{u} \cdot c\mathbf{w} = 0$

EDIT: Can we conclude this through some principle? Is it the transitive property?

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Did you intend to ask a question? –  Ittay Weiss Feb 3 '13 at 6:57
I would not say "by transitivity." –  André Nicolas Feb 3 '13 at 7:03
Has nothing to do with "transitive" property. There is no transitivity being used here. –  Thomas Andrews Feb 3 '13 at 7:05

We can certainly conclude it, but not by the transitive property.

The transitive property simply states that if $a = b$ and $b=c$, then $a=c$.

What we are using instead is the substitution property of equality: if $a=b$ and we have an expression involving $a$, we can replace $a$ with $b$ without affecting the expression. This substitution property is much stronger than transitivity.

EDIT: As an example, consider the relation $u \sim v$ when $\|u\| = \|v\|$. It's easy to see that $\sim$ is an equivelance relation, and in particular, satisfies the transitive property. But for arbitrary vectors $u,v,w$ with $\|v\|=\|w\|$, you can't go around replacing $v$ with $w$ in formulas like $u\cdot v$ and expect to get the same answer, namely, $u \cdot v = u \cdot w$.

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Transitivity only works for the same relation right? –  Hawk Feb 3 '13 at 7:10
In your example what do you mean by replace v with w in what formulae? Are you referring to the equality relation or the dot product? –  Hawk Feb 3 '13 at 7:20
Both? If $v$ and $w$ are equivalent under some transitive relation, that says nothing about whether $w$ can be substituted for $v$. Equality is special. –  user7530 Feb 3 '13 at 7:23
This is probably an absurd question. But assuming $u \cdot u = 1$, is doing $$u \cdot u \cdot v = u \cdot u \cdot w$$ illegal? I carried out this step after your example. I think it is wrong because your original equation is a scalar equality, but then I took the dot product of a scalar –  Hawk Feb 3 '13 at 7:26
Consider $u=(1,0,0)$, $v=(1,0,0)$, $w=(0,1,0)$. Then $v\sim w$, but $u\cdot v \neq u\cdot w$, and also $(u\cdot u) v \neq (u\cdot u) w$. –  user7530 Feb 3 '13 at 7:28