Vector Angles: Prove A.B = |A||B|cosADate: 07/13/98 at 02:37:31 From: Bryan J. Subject: Vector angles Hi Doctors, I just finished my pre-calc class, and at the end we covered vectors, including this theorem: A.B = |A||B|cosA dot product = length(length)cos(angle between them). If the proof isn't too complicated, what is the basis for that theorem? Any help would be appreciated. Bryan Date: 07/22/98 at 10:51:35 From: Doctor Rick Subject: Re: Vector angles Hi, Bryan. No, I don't think this proof is very complicated. Let's see if you agree. To start with, let's draw a vector A with length a, making an angle alpha with the x-axis: y | +....... + A | /: | / : | / : | a/ : | / : x = a*sin(alpha) | / : | / : |/)alpha : +--------+------x y = a*cos(alpha) The x component of the vector is a*cos(alpha), and the y component is a*sin(alpha), from the right triangle I have drawn. Now make another vector B, of length b and making angle beta with the x-axis: A = (a*cos(alpha), a*sin(alpha)) B = (b*cos(beta), b*sin(beta)) The dot product is now (just multiplying the x components and the y components, and adding them together) A . B = a*b*cos(alpha)*cos(beta) + a*b*sin(alpha)*sin(beta) = a * b * (cos(alpha)*cos(beta)+sin(alpha)*sin(beta)) All that's left is to remember the trigonometric identity, cos(alpha - beta) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta) and we have A . B = |A| * |B| * cos(alpha - beta) But (alpha - beta) is the angle between the two vectors. Does that explain it? Write back if you want more explanation. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 07/16/98 at 03:23:28 From: Bryan J. Subject: vector theorem proof Hi Doctors, In determing some angles in 3-D figures, I used the theorem: A.B = |A||B|cosA - pertaining to vectors I was thinking this looked like the law of cosines, but I couldn't maket the mental connection. Could you please help me to understand how the above theorem was derived? Thanks. Date: 07/16/98 at 07:08:18 From: Doctor Anthony Subject: Re: vector theorem proof This is not so much a 'theorem' as a 'definition'. You DEFINE the scalar products of two vectors to be the product: a.b = |a| x |b| x cos of angle between a and b. Following this definition you get the very convenient result that two vectors given in component form such as: a = (x1,y1,z1) b = (x2,y2,z2) then the scalar product is: a.b = x1.x2 + y1.y2 + z1.z2 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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