Homework 12.4-solutions

5 2 3 v 1 4 v 1 keywords vectors cross product

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Unformatted text preview: × P R| , ∆P QR has area = Now − − → PQ = 1− − →− → |P Q × P R | . 2 −5, 5, 0 , − → PR = |a × b| = |a| |b| sin θ where θ, 0 ≤ θ ≤ π , is the angle between a and b. Consequently, length = 24 sin(π ) = 0 . keywords: vector, cross product, length, 011 10.0 points Determine the scalar triple product, V , of the vectors a = i − j + 2 k, −2, 1, 0 . b = i − j + 2k , and But then i − − →− → P Q × P R = −5 −2 jk −5 5 k. 50= −2 1 10 1. V = −2 2. V = 0 correct Consequently, ∆P QR has area = c = −2 i − 2 j + k . 5 2 . 3. V = 1 4. V = −1 keywords: vectors, cross product area, triangle, parallelogram 010 10.0 points Determine the length of the cross product of a, b when |a| = 6, |b| = 4 and the angle between a, b is π . √ 1. length = 12 2 5. V = −3 Explanation: The scalar triple product is given by a · ( b × c) = = −1 −2 1 1 −2 1 2 + −2 1 −1 −1 −2 2 2 1 1 2 +2 −2 1 −1 . −2 mehmood (ajm4462) – Homework 12.4 – karakurt – (56295) Consequently, the scalar triple product of a, b and c is V = a · ( b × c) = 0 . 6 1. volume = 12 correct 2. vo...
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