# Pythagoras theorem in oblique coordinates

consider two oblique oblique basis vectors of unit length $$\vec{r_1}, \vec{r_2}$$

then any vector $$\vec{v} = p\vec{r_1}+q\vec{r_2}$$

define the dot product between two vectors a and b as $$|b|$$ (ie length of b) * (the length projection of(say) a on b made by an oblique line from a on b with angle same as that formed between $$\vec{r_1}, \vec{r_2}$$)

ie such that $$\vec{r_1} \cdot \vec{r_2} = 0$$

this is in contrast to the project of a on b along a right angle as done with orthogonal basis where you can use the cosine.

It's easy to show that this definition of the dot product is indeed distributive.

from our definition of v

$$\vec{v} \cdot \vec{v} = p^2 + q^2$$

Then $$|v|^2 = p^2 + q^2$$

which geometrically isn't the case,because p and q are oblique components, not perpendicular ones, this should not give you the length of v.

• You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles. – Yves Daoust Mar 18 at 16:01
• @YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above) – Vrisk Mar 18 at 16:08
• No, it doesn't have to. In oblique coordinates, this is false. – Yves Daoust Mar 18 at 16:24
• @YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length? – Vrisk Mar 18 at 16:29
• No, you get the scalar square of the vector. – Yves Daoust Mar 18 at 17:05

Let me expand on what Yves Daust is saying with a specific example.

Let $$\vec r_1 = (1,0)$$ and $$\vec r_2 =\left(\frac {\sqrt 2}2, \frac {\sqrt 2}2\right)$$ and suppose we consider the vector $$\vec v = \vec r_1 + \vec r_2$$.

By the definition of your inner product, $$|\vec v|^2 = 1^2 + 1^2 = 2$$.

Now if we express $$\vec v$$ in terms of canonical coordinates, it is $$\vec v = \vec r_1 + \vec r_2 = (1, 0) + \left(\frac {\sqrt 2}2, \frac {\sqrt 2}2\right) = \left(1+\frac {\sqrt 2}2, \frac {\sqrt 2}2\right)$$

And therefore its Euclidean norm is given by $$\|\vec v\|^2 = \left(1+\frac {\sqrt 2}2\right)^2 + \left(\frac{\sqrt 2}2\right)^2 = \left(1 + \sqrt 2 + \frac 12\right) + \frac12 = 2 + \sqrt 2$$

So very clearly, $$|\vec v| \ne \|\vec v\|$$.

$$|\vec v|$$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $$\Bbb R^2$$, so simply being a norm does not make it equal to that other more famous norm.

• Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ? – Vrisk Mar 19 at 14:37