Math 136 Winter 2011 Quiz 6
March 2, 2012
Name:
1. For each of the following questions use the vectors ~v =< 1, −3, 2 >, ~u =< −2, 1, 4 >, w
~ =< 1, 7 > and
m
~ =< −2, 5 >.
(a) Draw w
~ and m
~ on the same set of 2D axes. Then show using the geometry how to find m
~ +w
~
and 2m
~ − w.
~ Label the two new vectors clearly on your graph.
(b) Calculate 4~v − ~u.
4~v − ~u = 4 < 1, −3, 2 > − < −2, 1, 4 >=< 4, −12, 8 > − < −2, 1, 4 >=< 6, −13, 4 >
(c) Find a unit vector that points in the same direction as ~v .
First we compute the length or magnitude of ~v .
p
√
√
||~v || = 12 + (−3)2 + 22 = 1 + 9 + 4 = 14
A unit vector pointing in the same direction as ~v is then given by
1
1
~v = √ < 1, −3, 2 >=
||~v ||
14
1
1
v
||~
v || ~
or,
1
−3
2
√ ,√ ,√
14
14
14
2. Show that the equation 2x2 + 2y 2 + 2z 2 = 8x − 24z + 1 is an equation for a sphere and find the center
and radius of the sphere.
We begin by gathering the like terms:
2x2 − 8x + 2y 2 + 2z 2 + 24z = 1
Then we divided by 2 to obtain:
1
2
To complete the square for x we need to add in 4 and to complete the square in z we need to add in 36
giving:
(x2 − 4x + 4) + y 2 + (z 2 + 12z + 36) = 40.5
x2 − 4x + y 2 + z 2 + 12z =
Factoring then gives:
(x − 2)2 + y 2 + (z + 6)2 = 40.5
Thus this is an equation for a sphere that has its center at (2, 0, −6) and a radius of
√
40.5.
3. Answer #28 in section 13.2. Hint: Think about converting polar coordinates to cartesian coordinates
to write each of the vectors in standard form, then add the two vectors and convert the result back to
polar coordinates.
◦
The first vector that
√ is directed 45 above the x-axis with a magnitude of√20lb has an x (horizontal) component
of 20 cos(45) √
= 10 √
2. The y (vertical component) is 20 sin(45) = 10 2. Thus our first force, which I will
call F~1 = 10 2, 10 2 .
The second vector that is√directed 30◦ below the x-axis. Consequently the x (horizontal) component of the
◦
force is 16 cos(−30√
) = 8 3 and the vertical component is 16 sin(−30◦ ) = −8. Thus, the second force, which
~
i will call F2 = 8 3, −8 .
The resulting force for the two forces is the sum of F~1 and F~2 which is
D √
E
√
√
F~1 + F~2 = 10 2 + 8 3, 10 2 − 8 ≈ h27.9985, 6.1421i
The magnitude of the resulting force is just the length of this new vector which is:
r
2 p
√
√ 2 √
~
~
||F1 + F2 || =
10 2 + 8 3 + 10 2 − 8 ≈ 27.99852 + 6.14212 = 28.6643
The angle is clearly in the first quadrant since both coordinates are positive. Thus the angle can be found
by taking tan−1 (6.1421/27.9985) ≈ .21595 Radians ≈ 12.3731◦
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4. Read Section 13.3. Then given the vectors ~v =< 1, −3, 2 >, ~u =< −2, 1, 4 >, answer the following
questions.
(a) Compute ~v · ~u.
~v · ~u = (1)(−2) + (−3)(1) + (2)(4) = −2 − 3 + 8 = 3
(b) Find the angle formed between ~v and ~u.
To find the angle between vectors we use the identity:
cos(θ) =
~a · ~b
|~a||~b|
Using ~u and ~v instead of ~b and ~a we know that ~v · ~u = 3 from above. We now need the length of ~v and ~u.
We already did the length of ~v in problem 1 but we repeat it here:
p
√
|~v | = 12 + (−3)2 + 22 = 14
p
√
|~u| = (−2)2 + 12 + 42 = 21
Putting all of this together we then have,
cos(θ) = √
3
3
√ = √ ≈ 0.17496
14 21
7 6
Thus the angle between the two vectors is θ = cos−1 (.17496) = 1.3949 Radians.
(c) Find the scalar projection of ~v onto ~u.
The scalar projection of ~v onto ~u is given by:
comp~u~v =
3
~u · ~v
=√
|~u|
21
(d) Find the vector projection of ~v onto ~u.
The vector projection of ~v onto ~u is given by:
~u · ~v ~u
3
< −2, 1, 4 >
3
2 1 4
√
proj~u~v =
= √
=
< −2, 1, 4 >= − , ,
|~u| |u|
21
7 7 7
21
21
3
5. Consider the vectors ~v =< 4, −12, −8 >, ~u =< −3, 9, 6 >, and w
~ =< −3, −1, 0 >.
(a) Are the vectors ~v and ~u parallel? perpendicular? neither? (see page 817).
To check to see if the vectors are parallel or perpendicular we first take their dot product.
~v · ~u = (4)(−3) + (−12)(9) + (−8)(6) = −12 − 108 − 48 = −168
Since this is not 0, we can rule out that the vectors are perpendicular. Parallel vectors are scalar multiples
of each other. Is that the case here? In other words, is there a value c such that:
c < 4, −12, −8 >=< −3, 9, 6 >
If there was, the first component would tell us that 4c = −3 or c = −3/4, does that c work for the other
components as well? −12(−3/4) = 9 and −8(−3/4) = 6 so the answer is yes! Thus these two vectors are
scalar multiples of each other and hence are indeed parallel. Another way to check this is to use the angle
formula
cos(θ) =
−168
−168
~v · ~u
√
√
=√
= −1
=√
|~v ||~u|
16 + 144 + 64 9 + 81 + 36
224 126
Since cos(θ) = −1 then θ = π, which also indicates that the two vectors are parallel.
(b) Are the vectors ~v and w
~ parallel?, perpendicular? neither?
We again begin with the dot product of the two vectors,
~v · w
~ = (4)(−3) + (−12)(−1) + (−8)(0) = −12 + 12 + 0 = 0
Since the dot product is 0, then the two vectors are perpendicular.
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