Question

Decomposing a vector into an arbitrary basis with cramer's rule with geometric interpretation, all intuitively. @wio

Answer 1

Suppose we have a vector x in some arbitrary basis a and b, not necessarily orthogonal or normal.\[\large \bar x=\alpha \bar a+\beta \bar b\] We can take the wedge product of this to get \[\large \bar x \wedge \bar a = (\alpha \bar a+\beta \bar b) \wedge \bar a = \beta (\bar b \wedge \bar a)\]\[\large \bar x \wedge \bar b = (\alpha \bar a+\beta \bar b) \wedge \bar b = \alpha (\bar a \wedge \bar b)\] So now we can rewrite x in terms of wedge products: \[\large \bar x = \frac{\bar x \wedge \bar b}{\bar a \wedge \bar b} \bar a + \frac{\bar x \wedge \bar a}{\bar b \wedge \bar a}\bar b\] It might not seem intuitively obvious yet until you draw a picture, but remember the wedge product is essentially the area, which is just the determinant. I'l draw a picture in a second.

Answer 2

Th point is that the ratio of areas of x wedge a and a wedge b is the same amount you need to multiply b by to get the projection of x on the b-axis.

You can kind of see how x wedge a and a wedge b are dividing out a while "kind of leaving" x/b behind, allowing you to divide out the magnitude of b making it a unit vector if that makes sense.

Keep in mind that this is also cramer's rule. I'm sure I can go further in explaining it, but play with it a second because it's quite powerful once you see it.