It is “by definition”. Two non-zero vectors are claimed to be orthogonal once (if and also only if) their dot product is zero.

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Ok. But why did we define the orthogonality this way?


The algebraic definition

Yet, over there is also a geometric definition the the period product:

a b = ‖a‖ * ‖b‖ * cosøwhich is multiplying the size of the an initial vector through the length of the second vector v the cosine that the edge between the two vectors.

And the angle between the 2 perpendicular vectors is 90°.

When us substitute ø with 90° (cos 90°=0), `a•b` i do not care zero.

I want much more than simply a definition. Show me miscellaneous real?

Ok… then let me show you how those two meanings (geometric and also algebraic) agree v each other through the Pythagorean theorem.

Below is a simple recap the the vector norm. They will certainly be offered in expanding Pythagoras theorem come n-dimensions.


The size of n-dimensional vectors

Now, take into consideration two vectors <-1, 2> and <4, 2>.

Algebraically, <-1, 2> • <4, 2> = -4 + 4 = 0.

Applying the size formula ② come the Pythagorean theorem:



We can extend Pythagorean theorem to n-space.

Geometrically, you can see they are perpendicular together well.


Orthogonality displayed algebraically and also geometrically.

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The geometric an interpretation matches with the algebraic definition!

A pop Quiz:

Orthogonal vectors room linearly independent. This sound obvious- yet can friend prove that mathematically?

Hint: You deserve to use 2 definitions. 1) The algebraic definition of orthogonality2) The meaning of linear Independence: The vectors V1, V2, … , Vn space linearly independent if the equation a1 * V1 + a2 * V2 + … + an * Vn = 0 can only it is in satisfied by ai = 0 for every i.

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