벡터의 연산

The marble is subject to two forces. The first, gravity, is $-mg\,\hat{\mathbf{k}}$ with $m$ being the mass of the marble. The second is the normal force imposed by the plane. This forces acts in a direction perpendicular to the plane. One vector normal to the plane is $a\,\hat{\boldsymbol{\imath}}+b\,\hat{\boldsymbol{\jmath}}+c\,\hat{\mathbf{k}}$. So the force due to the plane is $T\left< a,b,c\right>$ with $T$ determined by the property that the net force perpendicular to the plane must be exactly zero, so that the marble remains on the plane, neither digging into nor flying off of it. The projection of the gravitational force onto the normal vector $\left< a,b,c\right>$ is

$$\frac{-mg\left< 0,0,1\right>\cdot\left< a,b,c\right>}{|\left< a,b,c\right>|^2}\left< a,b,c\right> =\frac{-mgc}{a^2+b^2+c^2}\left< a,b,c\right>$$

The condition that determines $T$ is thus

$$T\left< a,b,c\right>+\frac{-mgc}{a^2+b^2+c^2}\left< a,b,c\right> =0 \implies T=\frac{mgc}{a^2+b^2+c^2}$$

The total force on the marble is then (ignoring friction – which will have no effect on the direction of motion)

$$\begin{aligned} T\left< a,b,c\right>-mg\left< 0,0,1\right>&=\frac{mgc}{a^2+b^2+c^2}\left< a,b,c\right>-mg\left< 0,0,1\right>\\ &=mg\frac{c\left< a,b,c\right>-\left< 0,0,a^2+b^2+c^2\right>}{a^2+b^2+c^2}\\ &=mg\frac{\left< ac,bc,-a^2-b^2\right>}{a^2+b^2+c^2}\\ \end{aligned}$$

The direction of motion $\left< ac,bc,-a^2-b^2\right>$. If you want to turn this into a unit vector, just divide by $\sqrt{(a^2+b^2)(a^2+b^2+c^2)}$. Note that the direction vector in perpendicular $\left< a,b,c\right>$ and hence is parallel to the plane. If $c=0$, the plane is vertical. In this case, the marble doesn't roll – it falls straight down. If $a=b=0$, the plane is horizontal. In this case, the marble doesn't roll — it remains stationary.