HW21 - Sect. 12.4-solutions

# For the three given expressions therefore we see that

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Unformatted text preview: vectors, and its value is a vector; on the other hand, the dot product is deﬁned only for two vectors, and its value is a scalar. For the three given expressions, therefore, we see that I is well-deﬁned because both terms in the dot product are cross products, hence vectors. II is well-deﬁned because it is the dot product of two vectors. III is not well-deﬁned because the ﬁrst term in the cross product is a scalar because it is the length of a vector, not a vector. keywords: vectors, dot product, cross product, T/F, length, 004 2 10.0 points v=± a×b . |a × b| But for the given vectors a and b, i 1 2 a×b = = jk 43 10 9 1 43 i− 2 10 9 1 3 j+ 2 9 4 k 10 = 6i −3j + 2k. In this case, |a × b|2 = 49 . Consequently, v=± 3 2 6 i− j+ k 7 7 7 . Determine all unit vectors v orthogonal to a = i + 4 j + 3k, 1. v = ± b = 2 i + 10 j + 9 k . 3 6 2 i− j− k 7 7 7 2. v = 6 i − 3 j + 2 k 3. v = 6 2 3 i− j− k 7 7 7 4. v = 6 3 2 i− j− k 7 7 7 5. v = 3 i − 6 j − 2 k 6. v = ± 3 2 6 i − j + k correct 7 7 7 keywords: vector product, cross product, unit vector, orthogonal, 005 10.0 points If a is a vector parallel to the xy -plane and b is a vector parallel to k, determine |a × b| when |a| = 5 and |b| = 4. √ 1. |a × b| = 10 2 √ 2. |a × b| = −10 2 3. |a × b| = −20 4. |a × b| = 0 5. |a × b| = −10 cai (atc667) – HW21 - Sect. 12.4 – chavez-dominguez – (55235) 6. |a × b| = 20 correct is othogonal to the plane through P, Q and R. 7. |a × b| = 10 007 Explanation: For vectors a and b, |a × b| = |a||b| sin θ when the angle between them is θ , 0 ≤ θ < π . But θ = π/2 in the case when a is parallel to the xy -plane and b is parallel to k becau...
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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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