10c_04f_1bs

10c_04f_1bs - = 1 the area is | b | | c | sin θ 2 = √ 10...

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Math 10C First Midterm Version B Solutions Oct. 18, 2004 1. (a) No: You cannot add a vector and a scalar ( b · c is a scalar). (b) No: the cross product is only defned ±or vectors in R 3 . (c) Yes: this is a vector in R 3 . (d) No: a is in R 2 but v × w is in R 3 , so we cannot take their dot product. 2. (a) The answer is 3 a / | a | = (3 / 5 ) a = a 6 / 5 , 3 / 5 A . (b) We need a vector b = a b 1 , b 2 A so that 0 = a · b = 2 b 1 + b 2 . Thus b 2 = 2 b 1 . Any nonzero b 1 with b = a b 1 , 2 b 1 A is correct; ±or example, a 1 , 2 A = i 2 j . 3. Vectors ±or two sides o± the triangle are c = v AB = a− 1 , 3 , 0 A and b = v AC = a 2 , 0 , 1 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 2 0 1 1 3 0 v v v v v v = a− 3 , 1 , 6 A and the area is 3 2 + 1 2 + 6 2 / 2. You can leave the answer this way or simpli±y: 46 / 2. (ii) Since | b | = 5 , | c | = 10 and b · c = 2, it ±ollows that cos θ = 2 / 50. Thus, since sin 2 θ + cos 2 θ
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Unformatted text preview: = 1, the area is | b | | c | sin θ 2 = √ 10 √ 5 r 1 − 4 / 50 2 = √ 46 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 1 , 1 , 2 A · a x, y, z A = a 1 , 1 , 2 A · a 2 , − 1 , 1 A and so we have x + y + 2 z = 3. 5. We can use the ±ormula | ax + by + cz + d | |a a, b, c A| to obtain | 2 ∗ 1 + ( − 3) ∗ 2 + 1 ∗ 3 − 3 | r 2 2 + ( − 3) 2 + 1 2 = | − 4 | √ 14 = 4 √ 14 ....
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