Find a Point Above a SegmentDate: 02/16/99 at 03:56:17 From: Lebovitz Noam Subject: Multiplying vectors I have an assignment concerning multiplying vectors. I need to find a certain point above any segment. The point is set above the segment so it forms a triangle all of whose sides are equal (60 degrees for each angle in the triangle). The teacher said we should find the point by multiplying the segment vector by a vector that's 60 degrees from it, and whose quantity is 1, and that in order to multiply the vectors we should use complex numbers. I don't know how to do that. I even don't know how to represent a vector in a complex number form. This is an example of how the thing should look: the point ---> . The coordinates of the segment vector are given. the segment --> ___________ Thank you. Date: 02/16/99 at 07:53:25 From: Doctor Jerry Subject: Re: Multiplying vectors Hi Lebovitz, Thanks for your question. If the segment goes from P1 to P2, where P1 and P2 are represented by complex numbers, then you can calculate P2-P1, which will give you the complex number representation of the segment. The vector (expressed in complex numbers) that you want to multiply by is this one (it has unit length and has an angle of 60 deg, i.e. pi/3, from the horizontal axis): cos(pi/3) + i*sin(pi/3) As an example, suppose P1 is (1,1) or in complex form 1+i*1, P2 is (1+sqrt(3),2) or in complex form 1+sqrt(3)+i*2. P2-P1 = sqrt(3)+i. [cos(pi/3)+i*sin(pi/3)]*[P2-P1] = 2i. Note that the length of the segment from P1 to P2 is 2. The segment from P1 to P2 makes an angle of 30 degrees with the x-axis. Rotating this through 60 degrees makes it vertical. The length of the complex number cos(pi/3)+i*sin(pi/3) is 1, so that it doesn't change the length of the vector it is multiplying. In general, if you multiply the complex number a+b*i by cos(t)+i*sin(t), you will obtain (a*cos(t)-b*sin(t))+i*(b*cos(t)+a*sin(t)). These are strongly related to the rotation of coordinates equations. The rotation is through the angle t. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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