10c_04f_1as

# 10c_04f_1as - | b | | c | sin θ 2 = √ 10 √ 5 r 1 − 1...

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Math 10C First Midterm Version A Solutions Oct. 18, 2004 1. (a) No: a · w makes no sense since a R 3 and u R 2 . (b) Yes: this is a vector in R 3 . (c) No: the cross product is only defned ±or vectors in R 3 . (d) No: since ( a · b ) is a scalar, we cannot take its cross product with c . 2. (a) The answer is 2 a / | a | = (2 / 5 ) a = a 2 / 5 , 4 / 5 A . (b) We need a vector b = a b 1 , b 2 A so that 0 = a · b = b 1 + 2 b 2 . Thus b 1 = 2 b 2 . Any nonzero b 2 with b = a− 2 b 2 , b 2 A is correct; ±or example, a− 2 , 1 A = 2 i + j . 3. Vectors ±or two sides o± the triangle are c = v AB = a 1 , 2 , 0 A and b = v AC = a 1 , 0 , 3 A . Since the area is ( | b | | c | sin θ ) / 2, we can proceed in one o± two ways. (i) b × c = v v v v v v i j k 1 0 3 1 2 0 v v v v v v = a− 6 , 3 , 2 A and the area is 6 2 + 3 2 + 2 2 / 2. You can leave the answer this way or simpli±y: 49 / 2 = 7 / 2. (ii) Since | b | = 10 , | c | = 5 and b · c = 1, it ±ollows that cos θ = 1 / 50. Thus, since sin 2 θ + cos 2 θ = 1, the area is
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Unformatted text preview: | b | | c | sin θ 2 = √ 10 √ 5 r 1 − 1 / 50 2 = √ 49 2 = 7 / 2 . Remark : You may have chosen di²erent sides o± the triangle, di²erent directions on the sides, or di²erent order in the cross product. Nevertheless, you should get the same area and, to within sign, the same cross or dot product. 4. The equation is a 2 , − 1 , 1 A·a x, y, z A = a 2 , − 1 , 1 A·a 1 , 2 , 1 A and so we have 2 x − y + z = 1. 5. We can use the ±ormula | ax + by + cz + d | |a a, b, c A| to obtain | 3 ∗ 1 + 2 ∗ 2 + 1 ∗ ( − 3) − 2 | r 1 2 + 2 2 + ( − 3) 2 = 2 √ 14 ....
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## This note was uploaded on 06/17/2009 for the course MATH 3412341 taught by Professor Staff during the Fall '06 term at UCSD.

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