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Unformatted text preview: YOUR NAME: Alexander Givental Math 53. Midterm I. September 22, 2006. Theoretical question (15 pts) : Areas and 2 × 2-determinants. Given two vectors a = ( a 1 , a 2 ) and b = ( b 1 , b 2 ) in the plane, the determinant a 1 b 2- a 2 b 1 of the matrix formed by the coordinates of these vectors is equal to the signed area of the parallelogram P formed by these two vectors: det bracketleftbigg a 1 b 1 a 2 b 2 bracketrightbigg = ± Area ( P ) . The sign here is “+” is the direction of rotation from a to b is counter-clockwise, and “- ” is it is clockwise. Indeed, let φ and ψ be the angles that the vectors a and b respectively make with the positive direction of the 1st coordinate axis, and θ = ψ- φ be the angle of rotation from a to b (which is positive, when the direction of this rotation is counter-cklockwise). Then a 1 = | a | cos φ, a 2 = | a | sin φ, b 1 = | b | cos ψ, b 2 = | b | sin ψ, and hence a 1 b 2- a 2 b 1 = | a | | b | (cos φ sin ψ- sin φ cos ψ ) = | a | |...
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This note was uploaded on 03/09/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at Berkeley.
- Fall '08