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Unformatted text preview: 38) – HW 12.4 – radin – (56025) 4 Consequently, Consequently,
−
−
→−
→
v = P Q × P R = 20, 12, 15 volume = 48 . is othogonal to the plane through P, Q and
R.
008 10.0 points Compute the volume of the parallelopiped
with adjacent edges
PQ, PR, PS keywords: determinant, cross product, vector
product, scalar triple product, parallelopiped,
volume,
009 10.0 points determined by vertices
P (1, −1, 1) ,
R(3, −5, 3) , Find the maximum length of u × v when
u = 4 j and v is a position vector of length 7
in the yz plane. Q(3, −5, −3) ,
S (2, 1, 5) . 1. volume = 46 1. maximum length = 28 correct 2. volume = 47 2. maximum length = 26 3. volume = 49 3. maximum length = 24 4. volume = 48 correct 4. maximum length = 0 5. volume = 50 5. maximum length = 25 Explanation:
The parallelopiped is determined by the
vectors
−
−
→
a = P Q = 2, −4, −4 ,
−
→
b = PR =
−
→
c = PS = 2, −4, 2 , Thus its volume is given in terms of a scalar
triple product by
V =  a · ( b × c)  .
But
a · ( b × c) = 2 −4
1 Explanation:
The length of the cross product of u and v
is given by
u × v = u v sin θ = 28 sin θ 1, 2, 4 . 2 −4 6. maximum length = 27 −4 where 0 ≤ θ ≤ π is the angle between u
and v. Now j lies in the yz plane, so the
angle θ between j and v varies from 0 to π .
Consequently, u × v has
maximum length = 28 . 2 2 4
010 =2 −4
2 2
4 +4 2 2 1 4 −4 2
1 −4
2 . 10.0 points...
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 Spring '14

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