Length of vector perpendicular to plane

[GunnaSix]

Homework Statement

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?

Homework Equations

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?

 

[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.

 

[vela]

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 $\vec{c}$.

 

[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.