Resolving vectors into components parallel and perpendicular to a plane

Question

I have a 3D force vector at point x,y,z, and I have a vector A in a known direction at point x,y,z. I can determine the plane that this vector is normal to. I find the component of the force vector along vector A, which is in the direction perpendicular to the plane, by using the dot produce of the force vector and the vector A. Call this vec_perp. I then know the magnitude of the other vector by sqrt(force_vec^2 - perp_vec^2). I just want to clarify that the other vector will always lie in the plane defined by vector A which is normal to the plane and point x,y,z? Thanks

Answer 1

Suppose we have the force vector $f$ and another vector $n$ (which is not necessarily normalized, ie. does not have length 1).

The projection of $f$ onto $n$ (ie. the component of $f$ parallel to $n$) is defined to be $$f^{\Vert n} = \Vert f \Vert \dfrac{\langle f,n\rangle}{\Vert f\Vert\Vert n \Vert}\dfrac{n}{\Vert n\Vert}=\dfrac{\langle f, n\rangle}{\Vert n \Vert^2} \cdot n.$$

We claim that the perpendicular component of $f$ is given by $$f^{\perp n} = f-f^{\Vert n}.$$

Indeed the calculation $\begin{align} \langle f^{\perp n},f^{\Vert n}\rangle &= \langle f-f^{\Vert n},f^{\Vert n}\rangle\\ &= \langle f,f^{\Vert n} \rangle - \langle f^{\Vert n}, f^{\Vert n}\rangle\\ &= \langle f,\dfrac{\langle f,n\rangle}{\Vert n\Vert^2}\cdot n\rangle - \langle \dfrac{\langle f,n\rangle}{\Vert n \Vert^2}\cdot n, \dfrac{\langle f,n\rangle}{\Vert n \Vert^2}\cdot n\rangle\\ &=\dfrac{\langle f,n\rangle^2}{\Vert n \Vert^2} - \dfrac{\langle f,n\rangle^2}{\Vert n \Vert^4}\langle n,n\rangle\\ &=0 \end{align}$ demonstrates exactly that.