HW 12.4-solutions

# V b 9 i 10 j 2 k 3 6 2 i j k correct 7 7 7 4

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Unformatted text preview: 7 7 . Determine all unit vectors v orthogonal to a = −3 i + 4 j + k, 1. v = ± b = −9 i + 10 j + 2 k . 3 6 2 i + j − k correct 7 7 7 4. v = 6 3 2 i− j+ k 7 7 7 Find the area of the triangle having vertices 4. area = 7 5. area = 8 correct Explanation: The non-zero vectors orthogonal to a and b are all of the form v = λ(a × b) , λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus a×b . |a × b| Explanation: To use vectors we shall identify a line segment with the corresponding directed line segment. Since the area of the parallelogram having adjacent edges P Q and P R is given by − − →− → |P Q × P R| , ∆P QR has area = But for the given vectors a and b, = Now − − → PQ = jk 41 10 2 −3 1 i− −9 2 4 10 R(−1, −3) . 3. area = 9 6. v = −2 i − 3 j + 6 k a×b = Q(−2, 2) , 15 2 17 2. area = 2 3 6 2 5. v = − i − j + k 7 7 7 i −3 −9 10.0 points 1. area = 2 6 3 i− j+ k 7 7 7 v=± 004 P (1, 3) , 2. v = 2 i − 6 j + 3 k 3. v = ± keywords: vector product, cross product, unit vector, orthogonal, 1− − →− → |P Q × P R| . 2 −3, −1, 0 , − → PR = −2, −6, 0 . But then −3 1 j+ −9 2 = −2 i − 3 j + 6 k . In this case, |a × b|2 = 49 . 4 k 10 i − − →− → P Q × P R = −3 −2 j −1 −6 k −3 0= −2 0 Consequently, ∆P QR has area = 8 . −1 k. −6 morrell (mm59638) – HW 12.4 – radin – (56025) keywords: vectors, c...
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## This document was uploaded on 04/04/2014.

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