HW09-solutions

# Z 1 a b 11i 6j 9k geyer dag2798 hw09 rusin 55735 2 c

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Unformatted text preview: . 003 10.0 points Find the cross product of the vectors a = i − 2j − 3k , b = −3i − 2j + 3k . z 1. a × b = −11i + 6j − 9k geyer (dag2798) – HW09 – rusin – (55735) 2 C. a × b + b × a = 0. 2. a × b = −11i + 6j − 8k 3. a × b = −12i + 6j − 8k correct 1. A only 4. a × b = −12i + 4j − 8k 2. C only 5. a × b = −12i + 3j − 9k 3. A and C only correct 6. a × b = −11i + 4j − 9k 4. all of them Explanation: One way of computing the cross product (i − 2j − 3k) × (−3i − 2j + 3k) 6. A and B only 7. none of them is to use the fact that i ×j = k, 5. B only j× k = i, k ×i = j, 8. B and C only Explanation: while i×i = 0, j ×j = 0, k× k = 0. A. TRUE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then For then |a × b| = |a||b| sin θ , a × b = −12i + 6j − 8k . so if a = 0 and b = 0, then |a × b| = 0 Alternatively, we can use the deﬁnition a×b = = i 1 −3 −2 −2 j −2 −2 =⇒ sin θ = 0 . Thus θ = 0, π . In this case, a is parallel to b. k −3 3 B. FALSE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then 1 −3 i− −3 3 |a · b| = |a||b| cos θ , −3 j 3 while + 1 −3 −2 k −2 to determine a × b. 004 a × b = |a||b| sin θ . Thus |a · b|2 − |a × b|2 10.0 points Which of the following statements are true for all vectors a, b = 0? A. if a × b = 0, then a b, B. |a · b|2 − |a × b|2 = |a|2 |b|2 , = |a|2 |b|2 (cos2 θ − sin2 θ ) = |a|2 |b|2 cos 2θ = |a|2 |b|2 only when |a| = 0 or |b| = 0...
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