HW 12.4-solutions

# 0 points determined by vertices p 1 1 1 r3 5 3 find

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Unformatted text preview: 38) – HW 12.4 – radin – (56025) 4 Consequently, Consequently, − − →− → v = P Q × P R = 20, 12, 15 volume = 48 . is othogonal to the plane through P, Q and R. 008 10.0 points Compute the volume of the parallelopiped with adjacent edges PQ, PR, PS keywords: determinant, cross product, vector product, scalar triple product, parallelopiped, volume, 009 10.0 points determined by vertices P (1, −1, 1) , R(3, −5, 3) , Find the maximum length of u × v when u = 4 j and v is a position vector of length 7 in the yz -plane. Q(3, −5, −3) , S (2, 1, 5) . 1. volume = 46 1. maximum length = 28 correct 2. volume = 47 2. maximum length = 26 3. volume = 49 3. maximum length = 24 4. volume = 48 correct 4. maximum length = 0 5. volume = 50 5. maximum length = 25 Explanation: The parallelopiped is determined by the vectors − − → a = P Q = 2, −4, −4 , − → b = PR = − → c = PS = 2, −4, 2 , Thus its volume is given in terms of a scalar triple product by V = | a · ( b × c) | . But a · ( b × c) = 2 −4 1 Explanation: The length of the cross product of u and v is given by |u × v| = |u| |v| sin θ = 28 sin θ 1, 2, 4 . 2 −4 6. maximum length = 27 −4 where 0 ≤ θ ≤ π is the angle between u and v. Now j lies in the yz -plane, so the angle θ between j and v varies from 0 to π . Consequently, u × v has maximum length = 28 . 2 2 4 010 =2 −4 2 2 4 +4 2 2 1 4 −4 2 1 −4 2 . 10.0 points...
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