Why the inner product of two orthogonal vectors is zero

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SUMMARY

The inner product of two orthogonal vectors is always zero due to the definition of the inner product in real vector spaces, specifically R^n. The inner product is calculated as ||a|| * ||b|| * cos(theta), where theta is the angle between the vectors. For orthogonal vectors, cos(theta) equals zero, resulting in a zero inner product. This principle extends to any Euclidean space, as orthogonality implies that the vectors do not extend in each other's directions, confirming that their scalar product, defined as u^t · v, is zero.

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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?
 
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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 (<br /> \begin{array}{c}<br /> v_1\\ v_2 \\ \vdots \\ v_n<br /> \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 (<br /> \begin{array}{c}<br /> 0\\ 0 \\ \vdots \\ 1<br /> \end{array}\right )

see?
 
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Thank you, that's the kind of answer i was looking for :)
 
No worries - I edited to add another kind of answer while you were replying.
 

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