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Unformatted text preview: e product, parallelopiped, volume,
014 10.0 points Find the maximum length of u × v when
u = 2 j and v is a position vector of length 5
in the zxplane.
1. maximum length = 10 correct
2. maximum length = 11
3. maximum length = 12
4. maximum length = 0 and
c = 1, 1, 2 . 5. maximum length = 9 mehmood (ajm4462) – Homework 12.4 – karakurt – (56295)
6. maximum length = 13 6. d = Explanation:
The length of the cross product of u and v
is given by
u × v = u v sin θ = 10 sin θ
where 0 ≤ θ ≤ π is the angle between u and
v. Now j is perpendicular to the zxplane,
so the angle θ between j and v is always π/2.
Consequently, u × v has
maximum length = 10
015 . Explanation:
Graphically, d is the length of the perpen−→
−
dicular P D from P to ℓ shown in the ﬁgure.
Now by right angle trigonometry,
d = b sin θ .
On the other hand,
a × b = a b sin θ ;
i.e., 10.0 points
But then d= a
d 016 θ
P  a
1. d =
a·b −
−
→
b = QP . 10.0 points When P is a point not on the plane passing
thro...
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 Spring '10
 odell

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