View Full Version : Why the inner product of two orthogonal vectors is zero
flyerpower
Dec11-11, 01:26 PM
Why is the inner product of two orthogonal vectors always zero?
For example, in the real vector space R^n, the inner product is defined as ||a|| * ||b|| * cos(theta), and if they are orthogonal, cos(theta) is zero.
I can understand that, but how does this extend to any euclidean space?
Simon Bridge
Dec11-11, 01:34 PM
The inner product is just the amount that one vector extends in the direction of the other - it's how far one arrow "leans over" the other. If they are orthogonal, then they don't lean over each other at all. That's what "orthogonal means - they have zero extent in each other's directions.
This generalizes to many dimensions because each vector defines a direction in that space and the other one may have some lean in that direction.
Algebraically, if u and v are vectors, then their scalar product is defined as u^t \cdot v - in terms of components in an n-D basis that would be:
\left ( u_1,u_2, \ldots ,u_n \right ) \left (
\begin{array}{c}
v_1\\ v_2 \\ \vdots \\ v_n
\end{array}\right )
Try it for any two orthogonal vectors ... it's obvious for any two basis vectors...
\left ( 1,0, \ldots , 0 \right ) \left (
\begin{array}{c}
0\\ 0 \\ \vdots \\ 1
\end{array}\right )
see?
flyerpower
Dec11-11, 01:45 PM
Thank you, that's the kind of answer i was looking for :)
Simon Bridge
Dec11-11, 01:53 PM
No worries - I edited to add another kind of answer while you were replying.
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