HW21 - Sect. 12.4-solutions

# 1 volume 36 correct 2 volume 40 keywords cross product

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Unformatted text preview: se k is then perpendicular to the xy -plane. Consequently, for the given vectors, |a × b| = 20 . 3 10.0 points Compute the volume of the parallelopiped determined by the vectors a = 3, −2, 1 , b= 3, 4, 2 , and c= 3, 4, 4 . 1. volume = 36 correct 2. volume = 40 keywords: cross product, length, angle, 006 10.0 points Find a vector v orthogonal to the plane through the points P (2, 0, 0), Q(0, 5, 0), R(0, 0, 3) . 1. v = 5, 6, 10 3. volume = 38 4. volume = 37 5. volume = 39 Explanation: For the parallelopiped determined by vectors a, b, and c its volume = |a · (b × c)| . 2. v = 15, 6, 10 correct But 3. v = 15, 3, 10 4. v = 15, 2, 10 a · ( b × c) = 5. v = 3, 6, 10 Explanation: Because the plane through P , Q, R con− − → − → tains the vectors P Q and P R, any vector v orthogonal to both of these vectors (such as their cross product) must therefore be orthogonal to the plane. Here − − → P Q = −2, 5, 0 , − → P R = −2, 0, 3 . =3 42 44 3 −2 1 3 4 2 3 4 4 +2 32 34 + 3 4 3 4 . Consequently, volume = 36 . Consequently, − − →− → v = P Q × P R = 15, 6, 10 keywords: determinant, cross product, scalar triple product, parallelopiped, volume, cai (atc667) – HW21 - Sect. 12.4 – chavez-dominguez – (55235) 008 1. d = |a × b| |b| | a| a·b | a| |a × b| 6. d = 2. maximum length = 18 a·b | a| 5. d = 1. maximum length = 0 2. d = 4. d = Find the maximum length of u × v when u = 5 j and v is a position vector of length 3 in the xy -plane. |b| |a × b| 3. d = 10.0 points |a × b| correct | a| 3. maximum length = 15 correct 4. maximum length = 14 5. maximum length = 17 6. maximum length = 16 Explanation: The length of the cross product of u and v is given by Explanation: Graphic...
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## This note was uploaded on 02/20/2013 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas.

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