33_hwc1SolnsODDA

# 33_hwc1SolnsODDA - x y in IR3 that are orthogonal to the...

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Unformatted text preview: x y in IR3 that are orthogonal to the ﬁxed vector z Find an equation specifying this plane. 4.11 The set of all vectors 1 2 −1 is a plane in IR3 . x 1 SOLUTION Computing the dot product between the vectors x = y and 2 you z −1 will obtain the equation x + 2y − z = 0. That is the equation for a plane in IR3 , and x is 1 orthogonal to 2 if and only if x belongs to this plane. −1 4.13 Let {x1 , x2 , . . . , xk } be a collection of k orthogonal vectors in IRn , and let x = x1 + x2 + · · · + xk be their sum. Show that |x|2 = |x1 |2 + |x2 |2 + · · · + |xk |2 . This generalizes the Pythagorean Theorem to higher dimensions. SOLUTION Expanding one of the factors x in |x|2 = x · x, and using the distributive property of the dot product, we have |x|2 = (x1 + x2 + · · · + xk ) · x = x1 · x + x2 · x + · · · + xk · x . Now we claim that for each j = 1, . . . , k , xj · x = xj · xj = |xj |2 . (∗∗) (∗) Accepting the validity of (∗∗) for the moment, let’s see what it does for us: Using (∗∗) in each term in the ﬁnal sum in (∗), we have |x|2 = |x1 |2 + |x2 |2 + · · · + |xk |2 , which is just what we want. Therefore, we are done if we can verify (∗∗). But again using the distributive property of the dot product, xj · x = xj · (x1 + x2 + · · · + xk ) = xj · x1 + xj · x2 + · · · + xj · xk . By the orthogonality of the vectors, the only term in this sum that can be non zero is xj · xj = |xj |2 . Thus, (∗∗) is veriﬁed. 21/september/2005; 22:09 34 ...
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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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