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Calc Hw 8.pdf

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Department
Mathematics
Course
M 408C
Professor
Gary, Berg
Semester
Fall

Description
taboada (lat2278) – HW08 – berg – (56260) 1 This print-out should have 16 questions. Multiple-choice questions may continue on 1. length = 14 the next column or page – find all choices before answering. 2. length = 7 √ 3. length = 7 3 001 10.0 points Find the cross product of the vectors √ 4. length = 7 2 a = i − j + 3k, b = 2i + j + 2k. 5. length = 0 1. a × b = −4i + 2j + 2k 004 10.0 points 2. a × b = −4i + 4j + 2k If a is a vector parallel to the xy-plane and b is a vector parallel to k, determine |a × b| 3. a × b = −4i + 4j + 3k when |a| = 2 and |b| = 4. 4. a × b = −5i + 4j + 3k 1. |a × b| = −8 5. a × b = −5i + j + 2k 2. |a × b| = 4 6. a × b = −5i + 2j + 3k 3. |a × b| = 8 √ 002 10.0 points 4. |a × b| = 4 2 √ Determine all unit vectors v orthogonal to 5. |a × b| = −4 2 a = 4i + j − 3k, b = 14i + 3j − 12k. 6. |a × b| = 0 1. v = ± 6 i −3 j −2k 7. |a × b| = −4 7 7 7 005 10.0 points 3 6 2 2. v = − 7i + 7j − 7k Determine the scalar triple product, V , of 3 6 2 the vectors 3. v = ± i − j + k 7 7 7 a = −2i − j + 2k, b = 2i + 3j + k, 4. v = −3i + 6j − 2k and 5. v = −6i + 3j + 2k c = −2i + j + 2k. 6 3 2 6. v = − i + j + k 7 7 7 1. V = 13 003 10.0 points 2. V = 14 Determine the length of the cross product 3. V = 12 of a, b when |a| = 7, |b| = 2 and the angle between a, b is π/4. 4. V = 15 taboada (lat2278) – HW08 – berg – (56260) 2 5. V = 16 R ℓ 006 10.0 points D a Find a vector v orthogonal to the plane d through the points Q θ b P P(4,0,0), Q(0,3,0), R(0,0,5). Express the distance d from P to ℓ in terms of 1. v = ▯15,4,12▯ the vectors −→ −→ a = QR, b = QP . 2. v = ▯3,20,12▯ 3. v = ▯15,20,12▯ 1. d = |b| |a × b| 4. v = ▯5,20,12▯ |a| 2. d = |a × b| 5. v = ▯15,5,12▯ |a × b| 3. d = 007 10.0 points |a| 4. d = |a| a ▯ b Compute the volume of the parallelopiped determined by the vectors a ▯ b 5. d = |a| a = ▯4, −1, −4▯, b = ▯4, 3, 1▯, |a × b| 6. d = |b| and 009 10.0 points c = ▯3, 4, 1▯. When P is a point not on the plane passing through the points Q, R, and S, then the distance, d, from P to that plane is given by 1. volume = 33 the formula |a ▯ (b × c)| 2. volume = 29 d = |a × b| 3. volume = 32 where −→ −→ −→ a = QR, b = QS , c = QP . 4. volume = 31 Use this formula to determine the distance 5. volume = 30 from P(−2,2, −5) to the plane through the points 008 10.0 points Q(−2, −1, −1), R(−3, −1, 0), S(−3, 1, −1). Let ℓ be the line passing throug
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