TITLE: Equation for non-orthogonal projection of a point onto two vectors representing the isometric axis?
QUESTION [6 upvotes]: Suppose I have two vectors that are not orthogonal (let's say, an isometric grid) representing the new axis. Suppose I want to project a point onto these two vectors, how would I do it? Dot product only works for projecting a point orthogonally onto a vector, but now I want to project a point such that it is not along the orthogonal line, but respected to another vector.
For example, I found this picture in wikipedia: 
http://en.wikipedia.org/wiki/Projection_(linear_algebra)#Classification
Suppose I have two vectors that point in the same direction as line k and m. If I want to project a point onto axis m, I want to project it such that it follows along k onto m, not orthogonally. Then I can have my k and m as my new axis, and all the points will be projected respected to these two vectors.

REPLY [3 votes]: This projection onto two non-orthogonal vectors can be calculated more simply using the cross product. We know that the cross product of a vector with itself results in the zero vector. This is used to filter out both coefficients. 
Suppose we have three vectors $\textbf{v}_1$, $\textbf{v}_2$ and $\textbf{v}_3$ all lying in the plane defined by the perpendicular unit vector $\textbf{n}_z$. Now we wish to find the projection of vector $\textbf{v}_1$ onto $\textbf{v}_2$ and $\textbf{v}_3$ such that:
$$\textbf{v}_1=α\textbf{v}_2+β\textbf{v}_3$$
Using the cross product of both base vectors we get two equations:
$$\textbf{v}_2×\textbf{v}_1=α\textbf{v}_2×\textbf{v}_2+β\textbf{v}_2×\textbf{v}_3$$
$$\textbf{v}_3×\textbf{v}_1=α\textbf{v}_3×\textbf{v}_2+β\textbf{v}_3×\textbf{v}_3 $$
As the cross product with itself maps to zero we can write:
$$\textbf{v}_2×\textbf{v}_1=β\textbf{v}_2×\textbf{v}_3$$
$$\textbf{v}_3×\textbf{v}_1=α\textbf{v}_3×\textbf{v}_2  $$
Now the coefficients can be extracted for example using the fact the cross products are perpendicular to the plane defined by $\textbf{n}_z$. The dot product of the normal vector of the plane maps to the scalar:
$$\textbf{n}_z\cdot (\textbf{v}_2×\textbf{v}_1 )=β\textbf{n}_z\cdot (\textbf{v}_2×\textbf{v}_3)$$
$$\textbf{n}_z\cdot (\textbf{v}_3×\textbf{v}_1)=α\textbf{n}_z\cdot (\textbf{v}_3×\textbf{v}_2) $$
Resulting in the following system:
$$\alpha = \frac{\textbf{n}_z\cdot (\textbf{v}_3×\textbf{v}_1)}{\textbf{n}_z \cdot (\textbf{v}_3×\textbf{v}_2)}$$
$$ \beta = \frac{\textbf{n}_z\cdot (\textbf{v}_2×\textbf{v}_1)}{\textbf{n}_z \cdot (\textbf{v}_2×\textbf{v}_3)}$$
The nice thing is that this also work for non-unit length vectors and we need no matrix inversions.