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# PrQuiz3 - AMS210.01 Practice Quiz 3 Andrei Antonenko 1 Find...

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AMS210.01. Practice Quiz 3 Andrei Antonenko 1. Find the distance between the vector u = (10 , 5 , - 12) and the subspace, spanned by the following vectors: v 1 = (4 , 1 , 2); v 2 = (5 , 8 , 7); v 3 = (6 , - 3 , 0) . Solution: First we should apply the Gram-Schmidt orthogonalization process to vectors v 1 , v 2 , and v 3 : w 1 = v 1 = (4 , 1 , 2); w 2 = v 2 - h v 3 , w 1 i h w 1 , w 1 i w 1 = (5 , 8 , 7) - 20 + 8 + 14 16 + 1 + 4 (4 , 1 , 2) = (5 , 8 , 7) - 2(4 , 1 , 2) = ( - 3 , 6 , 3); w 3 = v 3 - h v 3 , w 1 i h w 1 , w 1 i w 1 - h v 3 , w 2 i h w 2 , w 2 i w 2 = (6 , - 3 , 0) - 24 - 3 16 + 1 + 4 (4 , 1 , 2) - - 18 - 18 9 + 36 + 9 ( - 3 , 6 , 3) = (6 , - 3 , 0) - (4 , 1 , 2) + 2 3 ( - 3 , 6 , 3) = (0 , 0 , 0) . So, we have: w 1 = (4 , 1 , 2); w 2 = ( - 3 , 6 , 3); w 3 = (0 , 0 , 0) . 1

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Now we can find the projection of u = (10 , 5 , - 12) along the subspace spanned by w 1 , w 2 , and w 3 . But w 3 = 0, so we simply omit it: proj( u ) = h u, w 1 i h w 1 , w 1 i w 1 + h u, w 2 i h w 2 , w 2 i w 2 = 40 + 5 - 24 16 + 1 + 4 (4 , 1 , 2) + - 30 + 30 - 36 9 + 36 + 9 ( - 3 , 6 , 3) = (4 , 1 , 2) - 2 3 ( - 3 , 6 , 3) = (6 , - 3 , 0) . Now, the orthogonal component of u is equal to u - proj( u ) = (10 , 5 , - 12) - (6 , - 3 , 0) = (4 , 8 , - 12) . Thus, the distance is equal to the norm of it: k (4 , 8 , - 12) k = 16 + 64 + 144 = 224 .
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