# Thread: Vector part of calculus course question

1. ## Vector part of calculus course question

Could someone please explain to me why the formula that is being used is being used in part c particularly (as in explaining the formula)?

To be clear, I am talking about the (r - r_0) * [(r_1 - r_0) x (r_2 - r_0)] = 0 formula. If you're unsure of what I am asking or need more information or something, just ask.

Sorry if this is not the right place to post but I don't know where else to post it since it's in my calculus course.

Any input would be appreciated!

2. ## Re: Vector part of calculus course question

I think I can.

First, the way I would do it. You have three points $P_0, P_1$ and $P_2$ all lying on the plane. If you create two vectors connecting two of the three points, say $\vec{P_0P_1}$ and $\vec{P_0P_2}$. If you cross these, this will give the normal to the plane

$\vec{n} = \vec{P_0P_1} \times \vec{P_0P_2}$.

Suppose that $(x,y,z)$ is a point lying on the plane then the vector throught this point and $P_0$ is

$$ (every line on the plane will contain this vector)

Now you want every line on the plane perpendicular to the normal so

$\vec{n} \cdot = 0.$

What I think the author is doing.

If you associate vectors for each point P_0, P_1 and P_2, from the origin to the point and denote these by ${\bf r}_0, {\bf r}_1$ and ${\bf r}_2$. Now

${\bf r}_0- {\bf r}_1 = \vec{P_0P_1}$ and ${\bf r}_0- {\bf r}_2 = \vec{P_0P_2}$

so

$\left({\bf r}_0- {\bf r}_1\right) \times \left({\bf r}_0- {\bf r}_2\right) = \vec{P_0P_1} \times \vec{P_0P_2} = \vec{n}$.

Now associate a vector from origin to $(x,y,z)$ as denote this as ${\bf r}$. Thus, the vector from $(x,y,z)$ to $P_0$ is

${\bf r} - {\bf r}_0 = $.

Now dot these two

$\left( {\bf r} - {\bf r}_0\right) \cdot \left({\bf r}_0- {\bf r}_1\right) \times \left({\bf r}_0- {\bf r}_2) = 0$

so you see the answers are the same.

Thanks!