34_hwc1SolnsODDA

34_hwc1SolnsODDA - so we have solved the problem with and...

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SECOND SOLUTION One can also use summation notation to do this, but one has to be careful with the indices: We have x = k j =1 x j . Therefore, | x | 2 = x · x = ± k X i =1 x i ! · k X j =1 x j . Note that we have used two different summation indices – i and j – for the two different sums, as we must. Now using the distributive property of the dot product, | x | 2 = k X i,j =1 x i · x j . ( * * * ) By the orthogonality, x i · x j = 0 unless i = j . Therefore, ( * * * ) reduces to | x | 2 = k X j =1 x j · x j = k X j =1 | x j | 2 . At the beginning, many students find arguments using summation notation a bit tricky, especially when it comes to when it matters how the indices are named. Though one can go quite far avoiding such arguments, it will eventually pay to become adept at them, and
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Unformatted text preview: so we have solved the problem with and without using summation notation. Who says you cant have it both ways? 4.15 Let u be the unit vector u = 1 5 h-4 3 i and x = h 5 7 i . Compute x k and x , where the direction is given by u . SOLUTION We use the formula x k = ( x u ) u to compute x k . Since x u =-20 + 21 5 = 1 5 , this gives x k = 1 25 -4 3 . Now, since x = x k + x , we have x = x-x k , which gives x = 5 7 -1 25 -4 3 = 1 25 25 35 --4 3 = 1 25 29 32 . 21 /september/ 2005; 22:09 35...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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