# Acceleration of a ball on a plane [closed]

I have a plane ($$ax+by+cz+d=0$$) in a 3D world, and a gravity vector $$\vec{g}$$ (say it's $$[0, 0, -9.81]$$.) How would I find the acceleration vector of an object on this plane, ignoring friction?

• If $\hat{n}$ is the unit vector normal to the plane (easily obtained from the plane equation), I think the answer is $$-\hat{n}\times(\hat{n}\times \vec{g}).$$ – Batominovski Apr 9 at 18:00

What you want to do is project $$\vec{g}$$ on the plane. First we will consider a new plane: $$ax+by+cz=0$$. This plane is parallel to our previous one, so our result won't change, but it goes through the origin. Then we need a normal vector of this plane. In this case, we're lucky because we have this particular form of the equation of the plane. A normal vector is in this case $$\vec{n}=[a,b,c]$$.

Now we will create a line going through $$\vec{g}$$, and parallel to $$\vec{n}$$. Then we get the following parameter-equation (I don't know the proper term): $$\vec{x}=\lambda\vec{n}+\vec{g}$$ Where $$\lambda$$ is the parameter. The intersection of this line and our plane will be $$\vec{g}$$ projected on our plane. Let's call this intersection-point $$\vec{p}$$. For $$\vec{p}$$ it must be true that: $$\vec{p}\text{ is on our line: }ap_x+bp_y+cp_z=\vec{p}\cdot\vec{n}=0$$ $$\vec{p}\text{ is on our plane: }\lambda\vec{n}+\vec{g}=\vec{p}$$ Plugging the second equation into the first we get: $$(\lambda\vec{n}+\vec{g})\cdot\vec{n}=\lambda\vec{n}\cdot\vec{n}+\vec{g}\cdot\vec{n}=0$$ $$\Rightarrow\lambda=-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}$$ Plugging this into the first equation gives us: $$\vec{p}=\vec{g}-\frac{\vec{g}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}\vec{n}$$ In your example ($$\vec{g}=[0, 0, -9.81]$$) it looks like: $$\vec{p}=\begin{bmatrix} \frac{9.81c}{a^2+b^2+c^2}a\\ \frac{9.81c}{a^2+b^2+c^2}b\\ \frac{9.81c}{a^2+b^2+c^2}c-9.81\\ \end{bmatrix}$$

• I think we got the same answer: $$-\hat{n}\times(\hat{n}\times \vec{g})=-(\hat{n}\cdot\vec{g})\hat{n}+\vec{g}.$$ Of course, $\hat{n}=\frac{\vec{n}}{\sqrt{\vec{n}\cdot\vec{n}}}$. – Batominovski Apr 9 at 18:06
• Nice, I always like it when there are multiple ways to get to the same answer – CodingDragon04 Apr 9 at 18:10
• I'm a little confused by your statement "...going through $\vec{g}$, and parallel to $\vec{n}$". I may be misunderstanding what you mean by "going through $\vec{g}$", but as I see it that would imply that the line has the same direction as $\vec{g}$, so how can it also be parallel to $\vec{n}$? – David Z Apr 10 at 1:52
• You should imagine $\vec{g}$ as being rooted at the origin. Then imagine a line in the direction of $\vec{n}$ and going through the tip of $\vec{g}$. Here I see $\vec{g}$ more as a point than a vector. – CodingDragon04 Apr 10 at 7:04

you are looking to find the component of the gravity vector in the direction of the line of greatest slope of the plane.

Taking the gravity vector as vertically downwards, the vector parallel to the plane which is horizontal is $$\underline{n}\times\underline{\hat{g}}=\left(\begin{matrix}-b\\a\\0\end{matrix}\right)$$

Then the line of greatest slope is $$\left(\begin{matrix}-b\\a\\0\end{matrix}\right)\times\underline{n}=\left(\begin{matrix}ac\\bc\\-a^2-b^2\end{matrix}\right)$$

Now calculate the component of $$\left(\begin{matrix}0\\0\\-g\end{matrix}\right)$$ in this direction, which is $$\frac{g(a^2+b^2)}{\sqrt{a^2c^2+b^2c^2+(a^2+b^2)^2}}$$

I like to think in terms of projections. The projection of a vector $$v$$ onto a plane with normal $$n$$ is $$p= v - n \;(v\cdot n) /(n \cdot n)$$ where $$(x_1, y_1, z_1) \cdot (x_2,y_2,z_2) = x_1 x_2 + y_1 y_2 + z_1 z_2$$.

So, the projection of the gravity vector onto your plane is just $$p = g - n\; (g\cdot n) /(n \cdot n).$$

Setting $$g=(0,0,-9)$$ and $$n=(a,b,c)$$, we get \begin{aligned} p &= (0,0,-9) - (a,b,c) \frac{(0,0,-9)\cdot (a,b,c)}{ a^2+b^2+c^2}\\ &= (0,0,-9) + (a,b,c)\frac{9c}{ a^2+b^2+c^2}. \end{aligned} (This matches the answers by WE-Tutotial-School, David Quinn, and CodingDragon04.)