How to find the projection vector of an arbitary vector on a plane?

Question

I have a 3D coordinate system defined by 3 perpendicular basis vectors (p), (q) and (r). On the other hand, I have an arbitary vector (d). I would like to find the vector (s), which is the projection of (d) on to one of the planes (w). p, q and r are normalized.

This is the visual representation of what I am trying to achieve:

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Could you please help me with this?

Answer 1

This means your plane is defined by the normalized vectors $\vec r$ and $\vec q$?. ($|\vec r|=1$ and $|\vec q|=1$) Then the projected vector $\vec s$ of $\vec d$ on this plane is simply:

$$\vec s = \left(\vec r\cdot \vec d\right)\vec r + \left(\vec q\cdot \vec d\right)\vec q$$