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M 408C
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Gary, Berg
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Mathematics

M 408C

Gary, Berg

Fall

Description

taboada (lat2278) – HW08 – berg – (56260) 1
This print-out should have 16 questions.
Multiple-choice questions may continue on 1. length = 14
the next column or page – ﬁnd all choices
before answering. 2. length = 7
√
3. length = 7 3
001 10.0 points
Find the cross product of the vectors √
4. length = 7 2
a = i − j + 3k, b = 2i + j + 2k.
5. length = 0
1. a × b = −4i + 2j + 2k
004 10.0 points
2. a × b = −4i + 4j + 2k If a is a vector parallel to the xy-plane and
b is a vector parallel to k, determine |a × b|
3. a × b = −4i + 4j + 3k when |a| = 2 and |b| = 4.
4. a × b = −5i + 4j + 3k 1. |a × b| = −8
5. a × b = −5i + j + 2k 2. |a × b| = 4
6. a × b = −5i + 2j + 3k 3. |a × b| = 8
√
002 10.0 points 4. |a × b| = 4 2
√
Determine all unit vectors v orthogonal to 5. |a × b| = −4 2
a = 4i + j − 3k, b = 14i + 3j − 12k. 6. |a × b| = 0
1. v = ± 6 i −3 j −2k 7. |a × b| = −4
7 7 7
005 10.0 points
3 6 2
2. v = − 7i + 7j − 7k Determine the scalar triple product, V , of
3 6 2 the vectors
3. v = ± i − j + k
7 7 7
a = −2i − j + 2k, b = 2i + 3j + k,
4. v = −3i + 6j − 2k
and
5. v = −6i + 3j + 2k c = −2i + j + 2k.
6 3 2
6. v = − i + j + k
7 7 7 1. V = 13
003 10.0 points
2. V = 14
Determine the length of the cross product 3. V = 12
of a, b when |a| = 7, |b| = 2 and the angle
between a, b is π/4. 4. V = 15 taboada (lat2278) – HW08 – berg – (56260) 2
5. V = 16
R ℓ
006 10.0 points D
a
Find a vector v orthogonal to the plane
d
through the points Q θ
b P
P(4,0,0), Q(0,3,0), R(0,0,5).
Express the distance d from P to ℓ in terms of
1. v = ▯15,4,12▯ the vectors
−→ −→
a = QR, b = QP .
2. v = ▯3,20,12▯
3. v = ▯15,20,12▯ 1. d = |b|
|a × b|
4. v = ▯5,20,12▯ |a|
2. d = |a × b|
5. v = ▯15,5,12▯
|a × b|
3. d =
007 10.0 points |a|
4. d = |a|
a ▯ b
Compute the volume of the parallelopiped
determined by the vectors a ▯ b
5. d = |a|
a = ▯4, −1, −4▯, b = ▯4, 3, 1▯, |a × b|
6. d =
|b|
and
009 10.0 points
c = ▯3, 4, 1▯.
When P is a point not on the plane passing
through the points Q, R, and S, then the
distance, d, from P to that plane is given by
1. volume = 33
the formula
|a ▯ (b × c)|
2. volume = 29 d = |a × b|
3. volume = 32 where
−→ −→ −→
a = QR, b = QS , c = QP .
4. volume = 31
Use this formula to determine the distance
5. volume = 30 from P(−2,2, −5) to the plane through the
points
008 10.0 points
Q(−2, −1, −1), R(−3, −1, 0), S(−3, 1, −1).
Let ℓ be the line passing throug

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