Intersecting Vectors and the Dot Product

Date: 04/24/98 at 03:05:47
From: Tony Lu
Subject: Geometry (vector)

Each of the following geometrical theorems can be proved by 
vectors using dot product(scalar product). Prove that: 

  i) the altitudes of a triangle are concurrent
 ii) the perpendicular bisectors of the sides of a triangle 
     are concurrent

I cannot figure out a way to prove that the three vectors intersect 
at a common point. I am not sure which properties of the scalar 
product to use. Can you please help me out?

Thanks.

Date: 04/24/98 at 08:30:00
From: Doctor Jerry
Subject: Re: Geometry(vector)

Hi Tony,

I'm not sure that what I'll suggest is the shortest, most elegant way, 
but I think it is correct. 

Let vertices of triangle be A, B, and C; let b be the vector from A to 
C, a the vector from C to B, and c be the vector from B to A. 

The altitude from A perpendicular to a can be written as:

     A:  b + L_A*a = h_A

where L_A (L sub A) is a scalar and h_A is a vector stretching from A 
to side BC. Similarly:

     B:  c + L_B*b = h_B

     C:  a + L_C*c = h_C

where h_A dot a = 0, h_B dot b = 0, and h_C dot c = 0.

First show that the line through A and parallel to h_A intersects the 
line through B and parallel to h_C intersect. Use A as origin for 
writing an equation for these lines:

     r = t*h_A

     r = b + s*h_C

where s and t are parameters. Set these equal, dot both sides with b, 
and solve for s:  (I'll use @ as dot)  

     s = -(a@b)/(a@h_C)

Now we must show the point b+s*h_C (with the above s) is on the line 
r=-c+w*h_B. Ask if the system:

     b+s*h_C = -c+w*h_B 

has a solution.

Dot both sides with c.

I think this works out.  Please check my work.

-Doctor Jerry,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/