Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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, $u \cdot v = u \cdot w$.

share|cite|improve this answer
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

Your Answer


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.