# Vector proof

• May 5th 2008, 04:40 AM
woollybull
Vector proof
Prove that, if (c-b).a = 0 and (c-a).b = 0, then (b-a).c=0. Show that this can be used to prove the following geometrical results.
(a) The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.

I can do the first part by using the distributive rule and communitative rule for scalar products.
But part (a) took a too much effort for what I think is meant to be a simple question. I showed that the lines through the vertices can be expressed in the one vector, therefore they must meet in a point. I used the fact that their dot product is zero. It took a hell of a lot of tedious algebra and I'm not even sure that its right. I'd appreciate any help.

Thanks.
• May 5th 2008, 07:27 AM
TheEmptySet
Quote:

Originally Posted by woollybull
Prove that, if (c-b).a = 0 and (c-a).b = 0, then (b-a).c=0. Show that this can be used to prove the following geometrical results.
(a) The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.

I can do the first part by using the distributive rule and communitative rule for scalar products.
But part (a) took a too much effort for what I think is meant to be a simple question. I showed that the lines through the vertices can be expressed in the one vector, therefore they must meet in a point. I used the fact that their dot product is zero. It took a hell of a lot of tedious algebra and I'm not even sure that its right. I'd appreciate any help.

Thanks.

$(\vec c -\vec b) \cdot \vec a=0$ and $(\vec c - \vec a)\cdot b =0$

Using the distributive law of the dot product we get

$\vec c \cdot \vec a -\vec b \cdot \vec a=0$ and $\vec c \cdot \vec b -\vec a \cdot \vec b =0$

using the communitive property of dot products we can rewrite the first equation as

$\vec c \cdot \vec a -\vec a \cdot \vec b=0$

Then multiply it by -1 and add it to the 2nd equation

$-\vec c \cdot \vec a +\vec a \cdot \vec b=0$
$\vec c \cdot \vec b -\vec a \cdot \vec b =0$

$\vec c \cdot \vec b - \vec c \cdot \vec a =0$

Now use the distribitive property in reverse(aka factor) out c

$\vec c \cdot (\vec b-\vec a)=0$

By the communitive property of the dot product we get

$(\vec b-\vec a) \cdot \vec c=0$

Yeah!! (Rock)
• May 6th 2008, 11:00 PM
woollybull
Thanks, but the initial part is fine, it is part (a) which I find to be very messy.
Does anyone have some advice for the second part of the question.