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# 20esolns1 - 20E HW 1 Solutions January 22, 2008 1.1.12)...

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Unformatted text preview: 20E HW 1 Solutions January 22, 2008 1.1.12) Answer P = {v1 + v2 : , R} 1.1.15) Answer Let v1 = -1i - 1j - 1k and v2 = 1i - 1j + 2k. Then L = {tv1 + (1 - t)v2 : t R} 1.1.17) Answer Let v1 = i + 3k and v2 = -2j. Then P = {v1 + v2 : 0 , , 1} 1.2.15) Answer proju (v) = uv -4 u= u uu 3 1.2.21) Answer Let p0 = 3i + 4j + 5k, and v = 400i + 500j - k. The airplanes position at time t is given by P (t) = p0 + tv. 1 The plane flies over the airport at time t satisfying: 3 + 400t = 23 4 + 500t = 29 Therefore at t = 1 of P ( 20 ) is 4.8. 1.3.1) Answer Note 1 20 (thus in 3 minutes) the plane is over the airport. The k (height) component 1 2 1 3 0 1 = 1 0 - 2 (6 - 2) + 1 0 = -8 2 0 2 3 0 1 1 2 1 = 3 (4 - 0) - 0 + 1 (0 - 4) = 8 2 0 2 and . 1.3.5) Answer i j k a b = 1 -2 1 = -3i + j + 5k 2 1 1 so Area = ||a b|| = 9 + 1 + 25 = 35. 1.3.7) Answer 2 1 -1 D = 5 0 -3 = -10 1 -2 1 so Area = |D| = 10. 1.3.16a) Answer v1 = 2i - k and v2 = 4j - 3k are vectors on the plane, so i j k v1 v2 = 2 0 -1 = 4i + 6j + 8k 0 4 -3 2 is normal to the plane. Thus the equation for the plane is given by: 4x + 6y + 8z = 0 (since (0,0,0) is on the plane). 3 ...
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## This note was uploaded on 04/07/2008 for the course MATH 20E taught by Professor Enright during the Fall '07 term at UCSD.

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20esolns1 - 20E HW 1 Solutions January 22, 2008 1.1.12)...

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