Intersecting Vectors and the Dot ProductDate: 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/ |
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