Mathematics 1229A/B Study Guide - Midterm Guide: Parallelepiped, Cross Product, Parallelogram
15 Oct 2018
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Friday, October 3, 2014
Page 1 CODE 111 Mathematics 1229A
Test 1
PART A (18 marks)
1. [1 mark ] Let u= (1,2,3), v= (2,1,3) and w= (3,2,1). Find the vector 2u−3v+w.
A: (11,10,16) B: (1,10,16) C: (1,3,2) D: (−1,3,−2) E: (−1,10,2)
Solution:
2u−3v+w= 2(1,2,3) −3(2,1,3) + (3,2,1) = (2,4,6) −(6,3,9) + (3,2,1)
= (2 −6 + 3,4−3 + 2,6−9 + 1) = (−1,3,−2)
2. [1 mark ] Find all values of kfor which the vector u= (k, 1) has kuk=√5.
A: 2 only B:√5C: 0 D: 5 E:±2
Solution: We know that
kuk=k(k, 1)k=pk2+ 12=pk2+ 1
so we need √k2+ 1 = √5 which means we need k2+ 1 = 5, i.e. k2= 4, which is true for both k= 2 and
k=−2.
3. [1 mark ] Find the vector u= 2(i−j+k) + 3(j−k), where i,jand kare unit vectors along the positive
x-axis, y-axis and z-axis, respectively.
A: (2,−1,1) B: (2,1,−1) C: (2,1,1) D: (0,1,−1) E: (1,−1,1)
Solution: Since we have k(and the zaxis) we are in ℜ3. In ℜ3we have i= (1,0,0), j= (0,1,0) and
k= (0,0,1). Therefore we get
u= 2(i−j+k) + 3(j−k)
= 2[(1,0,0) −(0,1,0) + (0,0,1)] + 3[(0,1,0) −(0,0,1)]
= 2(1,−1,1) + 3(0,1,−1) = (2,−2,2) + (0,3,−3) = (2,1,−1)
That is, we see that
u= 2(i−j+k) + 3(j−k) = 2i−2j+ 2k+ 3j−3k= 2i+j−k
and we know that the components of the vector uare just the multipliers on i,jand k, respectively, which
gives u= (2,1,−1).
4. [1 mark ] Find a unit vector in the direction opposite to u= (1,−1,1,−1,1).
A: (−1,1,−1,1,−1) B: (1,0,0,0,0) C: (−1,0,0,0,0)
D:1
√5,−1
√5,1
√5,−1
√5,1
√5E:−1
√5,1
√5,−1
√5,1
√5,−1
√5
Solution: A unit vector in the direction opposite to uis given by −1
cuwhere c=kuk. In this case we
have
c=kuk=k(1,−1,1,−1,1)k=p(1)2+ (−1)2+ (1)2+ (−1)2+ (1)2
=√1 + 1 + 1 + 1 + 1 = √5
which gives the vector we need as
−1
cu=−1
√5(1,−1,1,−1,1) = −1
√5,1
√5,−1
√5,1
√5,−1
√5
Mathematics 1229A
Test 1 CODE 111 Friday, October 3, 2014
Page 2
5. [1 mark ] Consider the points P(1,1), Q(2,3), R(4,0) and S(2,−4). Let ube the vector equivalent to
−→
P Q and vbe the vector equivalent to −→
RS. Which one of the following statements is true?
A:uand vare equal vectors. B:uand vare collinear but not equal. C:uand vare not collinear.
Solution: We get
u=−→
P Q =q−p= (2,3) −(1,1) = (1,2)
and v=−→
RS =s−r= (2,−4) −(4,0) = (−2,−4)
We see that these are not the same vector, so they are not equal vectors. But since (−2,−4) = −2(1,2)
we see that vis a scalar multiple of u, which tells us that uand vare collinear.
6. [1 mark ] Find the value of kfor which the vectors v= (2,1, k) and w= (4,2,1) are orthogonal
(perpendicular).
A:−10 B: 0 C:1
2D: 2 E: 10
Solution: In order for vand wto be orthogonal it must be true that v·w= 0. We see that
v·w= (2,1, k)·(4,2,1) = 2(4) + 1(2) + k(1) = 8 + 2 + k= 10 + k
and so we need 10 + k= 0, which is only true when k=−10.
7. [1 mark ] Find the value of cos θ, where θis the angle between the vectors v= (0,−1,2) and
w= (2,−2,1).
A: 0 B:1
3√5
C:4
√3
D:4
3√5
E:2
√3√5
Solution: We know that when θis the angle between vand w, cos θ=v·w
kvkkwk. Here we have
v·w= (0,−1,2) ·(2,−2,1) = 0(2) + (−1)(−2) + (2)(1) = 0 + 2 + 2 = 4
with kvk=k(0,−1,2)k=p02+ (−1)2+ 22=√0 + 1 + 4 = √5
and kwk=k(2,−2,1)k=p22+ (−2)2+ 12=√4 + 4 + 1 = √9 = 3
so that cos θ=4
√5(3) =4
3√5
8. [1 mark ] Let u= (1,1,−1) and v= (3,2,1). Find u×v.
A: 4 B: (3,2,−1) C: (3,4,−1) D: (−3,4,1) E: (3,−4,−1)
Solution: Using any (correct) method of finding the cross product vector we get
u×v= (1,1,−1)×(3,2,1) = (1(1)−(2)(−1),(−1)(3)−1(1),1(2)−3(1)) = (1+2,−3−1,2−3) = (3,−4,−1)
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MATH 1229A/B Full Course Notes
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Document Summary
[ 1 mark ] let u = (1, 2, 3), v = (2, 1, 3) and w = (3, 2, 1). Find the vector 2u 3v + w. 2u 3v + w = 2(1, 2, 3) 3(2, 1, 3) + (3, 2, 1) = (2, 4, 6) (6, 3, 9) + (3, 2, 1) = (2 6 + 3, 4 3 + 2, 6 9 + 1) = ( 1, 3, 2) [ 1 mark ] find all values of k for which the vector u = (k, 1) has kuk = 5. [ 1 mark ] find the vector u = 2(i j + k) + 3(j k), where i, j and k are unit vectors along the positive x-axis, y-axis and z-axis, respectively. Solution: since we have k (and the z axis) we are in 3. In 3 we have i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1).
