Length of vector perpendicular to plane

1. Jun 26, 2012

GunnaSix

1. The problem statement, all variables and given/known data
Let $\vec{a},\vec{b},\vec{c}$ be three constant vectors drawn from the origin to the points $A,B,C$. What is the distance from the origin to the plane defined by the points $A,B,C$? What is the area of the triangle $ABC$?

2. Relevant equations

3. The attempt at a solution
Starting with the first part:

The distance is the length of a perpendicular vector from the origin to the plane. If that vector is $\vec{r}$, then
$$\vec{r} \cdot (\vec{a} - \vec{b}) = \vec{r} \cdot (\vec{b} - \vec{c}) = \vec{r} \cdot (\vec{c} - \vec{a}) = \vec{r} \cdot (\vec{a} - \vec{r}) = 0\\ \vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} = r^2$$
by perpendicularity.

I can also get
$$\vec{r} \times [(\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})] = 0$$
by a similar argument, but I don't think it really helps.

I'm stuck here. Is there another relationship that I'm missing?

2. Jun 27, 2012

Muphrid

There's a simple formula that if you can write the equation of the plane in the form of $ax + by + cz + d = 0$, then the distance from that plane to the origin is $d/\sqrt{a^2 + b^2 + c^2}$. To be honest, I'm not sure how to prove this using just regular 3d vector geometry. This would be an ideal situation to use homogeneous coordinates, but I expect that's a bit more advanced.

3. Jun 27, 2012

vela

Staff Emeritus
You said $\vec{r}\cdot\vec{a} = r^2$. If you used the unit vector $\hat{r}=\vec{r}/||\vec{r}||$ instead, you'd have $\hat{r}\cdot\vec{a} = r$. Can you think of a way to find the unit normal to the plane in terms of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

4. Jun 27, 2012

GunnaSix

The cross product of two vectors between points in the plane is normal to the plane, so we have
$$\hat{r} = \frac{(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})}{\|(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})\|}$$

$$(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \vec{a} \times \vec{c} - \vec{a} \times \vec{b} - \vec{b} \times \vec{c} + \vec{b} \times \vec{b} = \vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}$$

$$r = \vec{a} \cdot \hat{r} = \frac{\vec{a} \cdot (\vec{a} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{b})}{\|\vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}\|} = \frac{\vec{a} \cdot (\vec{c} \times \vec{b})}{\|\vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}\|}$$

Thanks for the help.