HW21 - Sect. 12.4-solutions

Consequently u v has maximum length 15 009 d b sin

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Unformatted text preview: ally, d is the length of the perpen−→ − dicular P D from P to shown in the figure. Now by right angle trigonometry, |u × v| = |u| |v| sin θ = 15 sin θ where 0 ≤ θ ≤ π is the angle between u and v. Now j lies in the xy -plane, so the angle θ between j and v varies from 0 to π . Consequently, u × v has maximum length = 15 009 . d = |b| sin θ . On the other hand, |a × b| = |a| |b| sin θ ; i.e., |b| sin θ = 10.0 points Let be the line passing through points Q, R and P a point not on as shown in 4 |a × b| . | a| But then d= |a × b| . | a| R keywords: distance, distance from line, cross product, vectors D a d Q 010 θ b P Express the distance d from P to in terms of the vectors − − → − − → a = QR , b = QP . 10.0 points 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 the formula d= | a · ( b × c) | |a × b| cai (atc667) – HW21 - Sect. 12.4 – chavez-dominguez – (55235) where 5 Consequently, P is at a − − → a = QR , − → b = QS , − − → c = QP . Use this formula to determine the distance from P (2 , 0, −2) to the plane through the points Q(2, −2, 1), R(1, −2, 2), S (3, 0, 1) . 1. distance = 3 2 2. distance = 4 3. distance = 5 3 4. distance = 9 2 5. distance = 8 correct 3 6. distance = 3 Explanation: For the given points, − − → a = QR = i + 2 j , while − → b = QS = − i + k , and − − → c = QP = 2 j − 3 k . In this case a×b = i 1 −1 jk 20 01 = 2i− j + 2k, in which case |a × b| = 22 + 1 + 22 = 3 . On the other hand, a · ( b × c) = 1 −1 0 2 0 2 0 1 −3 = −8 . distance = 8 |−8| = 3 3 from the plane through Q, R and S . keywords: distance from plane, distance, plane, scalar triple product, cross product,...
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