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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 = ab 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 = ab cos θ , −3
j
3
while + 1
−3 −2
k
−2 to determine a × b.
004 a × b = ab sin θ .
Thus
a · b2 − a × b2 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 · b2 − a × b2 = a2 b2 , = a2 b2 (cos2 θ − sin2 θ )
= a2 b2 cos 2θ = a2 b2
only when a = 0 or b = 0...
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 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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