# Components of a vector onto nonorthogonal, nonunit basis vectors

I am trying to find the expression for the components of a vector along any two basis vectors in the plane. I tried using the dot product but this does not give me the right length! It is simply the projection of a vector onto a basis vector, but not the actual length of the component!

I am used to using the dot product to obtain components but this does not appear to work for general basis vectors.

Thanks.

• Well, the dot product obviously won't work if the two basis vectors are not orthogonal. Think what would happen if you tried it for one of the two basis vectors - it would give you a non-zero component along the other vector! – almagest Sep 26 '15 at 19:53
• Are all three vectors in the Cartesian plane? I.e. are we sure that the vector is in the span of the two "basis vectors in the plane"? – Rory Daulton Sep 26 '15 at 20:05
• This math.stackexchange.com/questions/1446037/… will help. – Michael Hoppe Sep 27 '15 at 11:25

If you're only dealing with the plane, here's a way.

Let $(v_1, v_2)$ be a basis of the XY plane, and suppose you want the components of $w$. You're looking for two scalars, $\alpha$ and $\beta$ such that $w = \alpha v_1 + \beta v_2$.

Cross product of two vectors in the XY plane is a vector that only has the Z component, so in this case we can abuse the notation a bit by pretending that it gives us scalars.

So cross the equation above by $v_1$ and $v_2$ respectively, obtaining:

$v_1 \times w = \beta v_1 \times v_2$

$v_2 \times w = \alpha v_2 \times v_1$

using the fact that $v_1 \times v_1 = v_2 \times v_2 = 0$.

Thus $\alpha = \frac{v_2 \times w}{v_2 \times v_1}$ and $\beta = \frac{v_1 \times w}{v_1 \times v_2}$.