How to project vector onto a plane but not along plane normal?

Question

In 3d space, is there any way to project a vector onto a plane, but along the UP direction (0,1,0) instead of the plane normal? If so, how do I do that and what is it called?

Answer 1

If I understand you correctly, you want to take an arbitrary vector $\vec{v}$ and decompose it as $\vec{v} = \vec{w} + \alpha \vec{u}$, where $\vec{w}$ lies in some plane with normal $\vec{n}$ (i.e., $\vec{w} \cdot \vec{n} = 0$), and $\vec{u}$ is a known vector, not necessarily equal to $\vec{n}$.

Since by assumption we have $\vec{v} = \vec{w} + \alpha \vec{u}$ and $\vec{w} \cdot \vec{n} = 0$, taking the dot product of both sides with $\vec{n}$ yields $$ \vec{v} \cdot \vec{n} = \alpha \vec{u} \cdot \vec{n} \quad \Rightarrow \quad \alpha = \frac{\vec{v} \cdot \vec{n}}{\vec{u} \cdot \vec{n}}. $$ This then implies that the desired vector $\vec{w}$ will be $$ \boxed{\vec{w} = \vec{v} - \left(\frac{\vec{v} \cdot \vec{n}}{\vec{u} \cdot \vec{n}} \right)\vec{u}.} $$

Note that in the case where $\vec{u} = \vec{n}$, this reduces to the standard orthogonal projection equation. Note also that in the case where $\vec{u}\cdot\vec{n} = 0$, this equation doesn't make any sense; this corresponds to the case where you're projecting along an axis lying parallel to the chosen plane, and so the desired vector $\vec{w}$ will not exist unless $\vec{v}$ already lies in that plane.

Answer 2

A picture would help, but this is what I think you are asking. Given the plane $ax+by+cz = d$ and the point $(p,q,r)$, you want to move the point in the direction $(0,1,0)$ until it meets the plane.

Well that move will change only the value of $q$, To find the new value, solve $$ ap + b? + cr = d $$ for the value of $?$.

There will be no solution if $(0,1,0)$ is parallel to the plane (unless the point is in the plane to begin with).

In more generality, you are asking for the intersection of a line (through a given point in a given direction) with a plane.