parallel vectors,, the Gram Schmidt procedure constructs a new set of vectors {u

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parallel vectors,, the Gram Schmidt procedure constructs a new set of vectors {ui, u2,U3} where each vector u. is a unit vector, and these three new vectors are mutually orthogonal (perpendicular) to each other. This new set of vectors can then act as usual set(i, This paper walks through an example of Given a set of 3 of three non a frame of reference similar to the the procedure. (Jörgen Pederson Gr German mathematician.] procedure. Jörgen Pederson Gram (1850-1916) Danish actuary. Erhardt Schmidt (1876-1959) 1 The procedure Initialization step: Consider the set and the cross product of any pair is not the zero vector. Hence, these vectors are 2-(1,0 1), and v3 = (0,0,1). } where ui-2 1,1), It is easy to see that these three are not all unit vectors. The dot product of any pair is nonzero, not mutually orthogonal (perpendicular), and no vector is parallel to another. ediate. To do the Gram-Schmidt procedure, follow these next steps. The vectors u are interm 1. First write v, -. Then form the unit vector 2. Compute w2-v2-(2) 1. Then form the unit vector u2 =--u 2. 3. Compute u 3 t 3 (t 3 . th)ui (v3 . tig)u2 Then form the unit vector u3- u 3. 4. It is clear from the construction that th, u2 are unit vectors. Now compute the dot products 1, First write wi = ui. Then form the unit vector ti wi w3 u; . u2, . th3,u2 . u3 to show that these are mutually orthogonal to each other. 2 A coordinate example It is a fact that given a set of mutually orthogonal unit vectors, any other vector may be written as a linear combination of these. This means, given any vector there exist scalars ,2, cs so that u may be written It is also true that the c scalars may be computed easily. For example, given a vector v, the distributive property of the dot product gives

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above question is improper.

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