Mathematics 1229A Test 1 2016
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CODE 111
Mathematics 1229A
Test 1
Friday, October 14, 2016
PART A (18 marks)
1. [1 mark] Which one of the following is not a unit vector?
A: (0, 1) |
B: ( 1^2 , − 1^2) |
C: ( − 35 , 0, − 45 , 0) |
D: ( 13 , 13 , 13) |
E: (0, − 1, 0, 0, 0) |
Solution: A unit vector is a vector whose magnitude is 1. For the vector ( 13 , 13 , 13), the magnitude is
( 13 , 13 , 13) = 4 (13)2 + ( 13)2 + ( 13)2 = 4 = 439 1
All of the other vectors given as answer choices do have magnitude 1.
2. [1 mark] Find a unit vector in the same direction as u = (−3, − 1, 4).
A: ( − 3^26 , − 1^26 , 4^26) |
B: ( − 3^6 , − 1^6 , 4^6) |
C: ( − 13 , − 1, 14) |
D: ( 13 , 1, − 14) |
E: ( − 14 , − 112 , 13) |
Solution: We know that a unit vector in the same direction as u is obtained by multiplying u by the scalar whose value is one over the magnitude of u. In this case we have
|u| = |(−3, − 1, 4)| = ^(−3)2 + ( − 1)2 + (4)2 = ^9 + 1 + 16 = ^26
which gives a unit vector in the same direction as u as
( ) u = ( 1^26) ( −3, − 1, 4) = ( − 3^26 , − 1^26 , 4^26)
Use the following information for questions 3 to 6.
Let u = (2, 1, 6), v = (0, 1, 1) and w = (3, 0, −3).
3. [1 mark] Find u + 2v − 3w.
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Solution: As above, to ind a scalar multiple of a vector, we multiply each component by the scalar. And
to ind a sum or diference of vectors, we add or subtract corresponding components. Therefore we get
u + 2v − 3w = (2, 1, 6) + (0, 2, 2) − (9, 0, −9) = (2 + 0 − 9, 1 + 2 − 0, 6 + 2 − ( −9)) = ( − 7, 3, 17)
4. [1 mark] Calculate w (2u).
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Solution: For the dot product of 2 vectors, we sum the products of corresponding components:
w (2u) = (3, 0, −3) (4, 2, 12) = 3(4) + 0(2) + ( −3)(12) = 12 + 0 − 36 = −24
5. [1 mark] Find |u − v + w|.
A: ^21 |
B: ^29 |
C: 7 |
D: ^57 |
E: ^65 |
Solution:
u − v + w = (2, 1, 6) − (0, 1, 1) + (3, 0, −3) = (2 − 0 + 3, 1 − 1 + 0, 6 − 1 + ( −3)) = (5, 0, 2) so |u − v + w| = |(5, 0, 2)| = ^52 + 02 + 22 = ^25 + 0 + 4 = ^29
6. [1 mark] Find the volume of the parallelepiped determined by u, v and w.
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Solution: The volume of the parallelepiped determined by u, v and w can be found using the calculation |(u × v) w|. Using your favourite method of inding a cross product you should have got
u × v = (2, 1, 6) × (0, 1, 1) = (1 − 6, 0 − 2, 2 − 0) = ( − 5, − 2, 2)
so (u × v) w = ( − 5, − 2, 2) (3, 0, −3) = − 15 − 0 − 6 = − 21
which gives Volume = |(u × v) w| = | − 21| = 21
7. [1 mark] Determine which one of the following vectors is orthogonal to both (2, 0, 1) and ( − 1, 3, 1).
A: (−3, −3, 6) B: (1, 1, 2) C: (− 1, 0, 2) D: (−3, 3, 6) E: (6, −3, 3)
Solution: We know that the cross product of 2 vectors is orthogonal to both of the vectors, so we can start by calculating the cross product of the two given vectors:
(2, 0, 1) × ( − 1, 3, 1) = (0 − 3, − 1 − 2, 6 − 0) = ( −3, −3, 6)
We do see this vector among the answer choices.
8. [1 mark] Find the cosine of the angle between the vectors (1, − 1, 1, 1) and (2, 2, 0, − 1).
A: 16 |
B: − 16 |
C: 15 |
D: − 15 |
E: 136 |
Solution: We know that the cosine of the angle θ between two vectors is given by the dot product of the vectors divided by the product of the magnitudes of the vectors. In this case we get:
(1, − 1, 1, 1) (2, 2, 0, − 1) = 2 − 2 + 0 − 1 = − 1
with |(1, − 1, 1, 1)| = ^12 + ( − 1)2 + 12 + 12 = ^1 + 1 + 1 + 1 = ^4 = 2
and |(2, 2, 0, − 1)| = ^22 + 22 + 02 + ( − 1)2 = ^4 + 4 + 0 + 1 = ^9 = 3
= = = −
9. [1 mark] If u = 2i + 3j − 4k and v = i + j + k, ind u × v.
A: −i − 2j + 5k B: −i − 6j − k C: −i + 6j − k D: 7i + 6j − k E: 7i − 6j − k
Solution: Since we have k, we must be in ℜ3 . We know that i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1), and that a vector which is stated as a sum of multiples of these vectors has components given by the
corresponding multipliers. So we have u = (2, 3, −4) and v = (1, 1, 1). Therefore the cross product of
(2, 3, −4) × (1, 1, 1) = (3 − ( −4), −4 − 2, 2 − 3) = (7, −6, − 1) = 7i − 6j − k
10. [1 mark] Which one of the following lines below?
ℓ :
is parallel to the line ℓ with parametric equations shown
x = 3 + 2t
y = −t
z = 2 + 5t
A: (x,y,z) = (4 , − 2 , 10) + t(1 , 1 , 1) |
B: (x,y,z) = (2 , − 1 , 5) + t(1 , 1 , 1) |
C: (x,y,z) = (1 , 1 , 1) + t(4 , − 2 , 10) |
D: (x,y,z) = (3, 0, 2) + t(2, 0, 5) |
E: (x,y,z) = (2, 1, 5) + t(3, 0, 2) |
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Solution: The line ℓ is parallel to the vector whose components are the multipliers on t in the parametric
equations, in this case the vector (2, − 1, 5). The answer choices are all point-parallel form equations of
being a vector which is collinear with (2 , − 1 , 5), i.e. u must be a scalar multiple of (2 , − 1 , 5) (where of
course the scalar multiplier could be 1, but we don’t see an equation that has (2 , − 1 , 5) being multiplied
by the parameter). We see that (4 , − 2 , 10) = 2(2 , − 1 , 5), so this is another vector which is parallel to line
11. [1 mark] Which of the following is a two-point form equation of the line through the points (2, 2, 0, 1) and (1, −6, 3, − 1)?
A: (x1 ,x2 ,x3 ,x4 ) = (2 , 2 , 0 , 1) + t(1 , −6 , 3 , − 1) |
B: (x1 ,x2 ,x3 ,x4 ) = (1 + t)(1 , −6 , 3 , − 1) + t(2 , 2 , 0 , 1) |
C: (x1 ,x2 ,x3 ,x4 ) = s(2 , 2 , 0 , 1) + t(1 , −6 , 3 , − 1) |
D: (x1 ,x2 ,x3 ,x4 ) = (1 − t)(2 , 2 , 0 , 1) + t(1 , −6 , 3 , − 1) |
E: (1, 6, 3, 1) ((x1 ,x2 ,x3 ,x4 ) (2, 2, 0, 1)) = 0 |
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Solution: A two-point form equation of the line through points P and Q has the form x = (1 − t)p + tq.
Considering the points as P(2 , 2 , 0 , 1) and Q(1 , −6 , 3 , − 1) we see that the one of the answer choices which
(x1 ,x2 ,x3 ,x4 ) = (1 − t)(2, 2, 0, 1) + t(1, −6, 3, − 1)
12. [1 mark] Which of the following is a point-normal form equation of the plane through the point (1, 3, 4) with normal vector (1, 0, 5)?
A: (1 , 0 , 5) (x − (1 , 3 , 4)) = 0 |
B: (1 , 3 , 4) (x + (1 , 0 , 5)) = − 21 |
C: (1 , 3 , 4) (x − (1 , 0 , 5)) = 0 |
D: (1, 0, 5) (x − (1, 3, 4)) = 21 |
E: (1, 0, 5) (x + (1, 3, 4)) = 0 |
2023-06-08