Prove a vector is perpendicular to a plane?

[dangsy]

Homework Statement

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

Homework Equations

Cross Product

The Attempt at a Solution

I tried writing everything out doing the Det of each Cross product to find anything to cancel without 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 =)

 

[Kurdt]

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?

 

[dangsy]

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?

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 =(

 

[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.

 

[Leong]

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

[tex]
\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*}
[/tex]

This Latex is extremely difficult to use.