how to find a vector, perpendicular to the normal of a plane, for which gravity is the strongest?

Question

Suppose a plane is inclined to some degree in three dimensional space, with gravity acting opposite to the z axis of the world. How would you find a vector, that runs along the plane in the direction that gravity is the strongest.

For example, say you have a piece of glass, and you tilt it in some way and place a ball on its surface. How to find a vector that represents the direction which the ball would roll?

Answer 1

Let $\mathbf{F}$ be the gravitational force acting on the ball. You can split $\mathbf{F} = \mathbf{F}^{\parallel}+\mathbf{F}^\bot$ into a component $\mathbf{F}^{\parallel}$ parallel and a component $\mathbf{F}^\bot$ perpendicular to the plane's normal vector. The parallel component is countered by a force of equal amount into the opposite direction, preventing the ball from passing through the board. The perpendicular component is the force moving the ball forward.

Let $\mathbf{n}$ be the plane's normal vector of unit length. Then the product $\mathbf{n}\cdot\mathbf{n}^t$ is the projection matrix onto $\mathbf{n}$, where $\mathbf{n}^t$ denotes the transpose of $\mathbf{n}$ (laying the vector down into the horizontal).

You then get

$$ \mathbf{F}^\parallel = \mathbf{n}(\mathbf{n}^t\cdot\mathbf{F}) $$

from the projection onto the normal vector. And from the split

$$ \mathbf{F}^\bot = \mathbf{F} - \mathbf{F}^\parallel = \mathbf{F} - \mathbf{n}(\mathbf{n}^t\cdot\mathbf{F}) $$

To see that $\mathbf{F}^\bot$ is indeed perpendicular to the normal vector, calculate the scalar product with $\mathbf{n}$:

$$ \mathbf{n}^t \cdot \mathbf{F}^\bot = \mathbf{n}^t \cdot \mathbf{F} - (\underbrace{\mathbf{n}^t \cdot\mathbf{n}}_{=1})(\mathbf{n}^t \cdot \mathbf{F}) = 0 $$

Answer 2

Well, using vectors it becomes really easy. Let the plane have normal vector ${\hat n}$ and acceleration be along ${\vec g}$. You know that the required vector, ${\vec g}$ and ${\hat n}$ will lie along the same plane.

The vector along the direction the ball would roll is given by $${\vec a}={\hat n}\times({\vec g}\times{\hat n})$$ $${\hat a}={{\hat n}\times({\vec g}\times{\hat n})\over |{\vec g}|}$$