# Prove a vector is perpendicular to a plane?

1. Jan 16, 2008

### dangsy

1. The problem statement, all variables and given/known data
Vectors A, B and C are vectors from the origin to the points a, b, c respectively, and the Vector D is defined as

D= (AxB)+(BxC)+(CxA)

Show that D is Perpendicular to the plane in which the points A, B, and C lie

2. Relevant equations
Cross Product

3. The attempt at a solution
I tried writing everything out doing the Det of each Cross product to find anything to cancel with out any luck.

I also understand that if a b and c are points lying on a plane, then crossing each of these vectors AxB, BxC, and CxA should give me a vector perpendicular to that plane...how do I show this mathmatically?

Thanks for the help =)

2. Jan 16, 2008

### Kurdt

Staff Emeritus
Couple of things I suppose. The dot product of perpendicular vectors is zero. so D dotted into the vector from A to B or B to C or C to A etc. should all be zero. Do you know the vector equation of a plane?

Last edited: Jan 16, 2008
3. Jan 16, 2008

### dangsy

n . (r->p) = 0

if n is the normal vector ( D in this case I think )
r and p are the points in the plane ( A->B, B->C, C->A)

I'm still not understanding how it relates =(

4. Jan 17, 2008

### Shooting Star

Hi dangsy, a shortcut (if it's allowed by your teacher)!

Three points always lie in a plane -- that we know. Chose the point 'a' as the origin.

Then A = 0, which gives,

D = BXC, which we know is perp to the plane.

5. Jan 17, 2008

### Leong

Let $$\overrightarrow{n}$$= Normal vector of the plane where points a, b and c lie on.

$$\begin{equation*} \begin{split} \overrightarrow{n} &= \overrightarrow{ab}\ x\ \overrightarrow{bc} \\ &= (\overrightarrow{B} - \overrightarrow{A})\ x\ (\overrightarrow{C} - \overrightarrow{B}) \\ &= [(\overrightarrow{B} - \overrightarrow{A})\ x\ \overrightarrow{C}] - [(\overrightarrow{B} - \overrightarrow{A})\ x\ \vec{B}] \\ &= [(\overrightarrow{B} \ x\ \overrightarrow{C})\ - (\overrightarrow{A} \ x\ \overrightarrow{C})] \ -\ [(\overrightarrow{B} \ x\ \overrightarrow{B})\ - (\overrightarrow{A} \ x\ \overrightarrow{B})] \\ &= [(\overrightarrow{B} \ x\ \overrightarrow{C})\ + (\overrightarrow{C} \ x\ \overrightarrow{A})] \ +\ (\overrightarrow{A} \ x\ \overrightarrow{B})] \\ &=(\overrightarrow{A} \ x\ \overrightarrow{B})\ + (\overrightarrow{B} \ x\ \overrightarrow{C}) \ +\ (\overrightarrow{C} \ x\ \overrightarrow{A}) \ Note\ that: -(\overrightarrow{A} \ x\ \overrightarrow{C}) = \overrightarrow{C} \ x\ \overrightarrow{A}\ and \ \overrightarrow{B} \ x\ \overrightarrow{B} = \overrightarrow{0} \\ &=\overrightarrow{D} \\ &\overrightarrow{n} = \overrightarrow{D} \\ &\ which\ means\ \overrightarrow{n} \ is \ parallel \ with\ \overrightarrow{D} \\ &\overrightarrow{n}\ is \ perpendicular \ to \ the \ plane.\ Therefore,\ \overrightarrow{D} \ is\ also \ perpendicular\ to \ the \ plane. \end{split} \end{equation*}$$

This Latex is extremely difficult to use.

Last edited: Jan 17, 2008