hw1.2.34 - ±u and get proj ±u ±v = ±u ±v || ±u || 2...

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Math 43, Fall 2007 Week 2: Finish suggested homework 1, . . . Graded Homework 2: 1.2 - 34 and 1.3 - 4, due Mon. Sept 10 Suggested Homework 2: problems from graded Homework 2 plus 1.2 - 22, 23, 35, 36; 1.3 - 6, due (but not collected) Wed. Sept. 12.
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2 Second week. Problem 1.2.9, Section 1.2, pg. 26: Find || a || and a unit vector in the direction of a [above] [Recall the formula u = 1 || a || a. ] material finished from week 1: 5. linear combinations, standard coordinates and new coordinates.
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new material Problem 1.2.34, Section 1.2, pg. 26: Find the projection of v = [1 , - 1] onto u = [3 , - 1] . Solution: The projection onto u is usually pictured as a vector in the direction of u. Trig and the formula for cos in terms of the dot product tells us that the length of the projection is (usually) || proj u ( v ) || = u · v || u || . We multiply that number by the unit vector
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Unformatted text preview: ±u and get proj ±u ( ±v ) = ±u · ±v || ±u || 2 ±u. 4 For the unit vector in the direction of [3 ,-1] , we find || ±u || = √ 9 + 1 = √ 10 , and for the dot product we have ±u · ±v = [1 ,-1] · [3 ,-1] = 3 + 1 = 4 , so proj ±u ( ±v ) = ±u · ±v || ±u || 2 ±u = 4 10 ±u = 2 5 [3 ,-1] . For the start of Section 1.3, we use the vector form of the equation of a line ±x = ± p + t ± d, where ± p is a point on the line, ± d is the direction vector of the line, and ±x is in R 2 or R 3 . See the text Example 1.20, pg. 32-33, for discussion. When the point on the line is (1 , 3) and the direction is ± d = [1 ,-2] , we get ±x = [ x, y ] = [1 , 3] + t [1 ,-2] , which we write in terms of components as x = 1 + t, y = 3-2 t....
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