Math 217: Selected Solutions to Assignment 1
12.1 # 12: The sphere with centre (2, −6, 4) and radius 5 has equation
(x − 2)2 + (y + 6)2 + (z − 4)2 = 25.
Its intersection with the xy-plane (z = 0) is
(x − 2)2 + (y + 6)2 = 25 − 16 = 9
is the circle of radius 3 centred
√ Similarly, the intersection with the yz-plane
√ at (2, −6).
(x = 0) is a circle of radius 25 − 4 = 21. The intersection with the xz-plane (y = 0)
has equation
(x − 2)2 + (z − 4)2 = 25 − 36 = −11
which has no solutions – it is the empty set.
12.1 # 30: In 3-space, y 2 + z 2 = 16 is the cylinder, extending in the x-direction,
whose cross-section in the yz-plane is the circle of radius 16 centred at the origin.
12.1 # 34: complete the square in x2 + y 2 + z 2 − 2z ≥ 0 to get
x2 + y 2 + (z − 1)2 − 1 ≥ 0
−−>
x2 + y 2 + (z − 1)2 ≥ 1,
which is the exterior to the sphere of radius 1 centred at (0, 0, 1) (including the spree itself).
12.2 # 44: Draw a picture. Since C lies on AB, we must have that c−a is a multiple
of b − a. And since BC is twice the length of AC, we see that c − a is 1/3 of b − a. So
c − a = (1/3)(b − a), and some simple algebra yields c = (2/3)a + (1/3)b.
12.3 # 28: Let a = ha1 , a2 i be one of the unit vectors we are looking for. The angle
between v and a should be π/3:
1/2 = cos(π/3) =
v·a
3a1 + 4a2
=
|v||a|
5
(here we used |a| = 1). So 6a1 + 8a2 = 5, and so a2 = (5 − 6a1 )/8. Now substitute this
into the unit length condition a21 + a22 = 1 to get
1 = a21 + (5 − 6a1 )2 /64 = (1/64)(100a21 − 60a1 + 25)
1
so 100a21 − 60a1 − 39 = 0. The quadratic formula gives
√
√
(60)2 + (100)4(39)
3 ± 9 + 39
3±4 3
a1 =
=
=
.
200
10
10
√
Solving to get a2 = 5/8−(3/10)(3±4 3), we have the components of two vectors (choosing
either the plus sign or the minus sign) satisfying the question.
60 ±
p
12.3 # 56: Suppose the cube has side length a, and place it in the first octant (x > 0,
y > 0, z > 0) with a corner at the origin. Then the vector v1 = a < 1, 1, 1 > points in the
direction of the diagonal of the cube, and the vector v2 = a < 1, 1, 0 > points along the
diagonal of one of the sides. The angle between them is
θ = cos−1
v1 · v2
|v1 ||v2 |
= cos−1
2a2
√
√
( 3a)( 2a)
= cos−1
√ !
2
√
.
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12.5 # 62: The plane of points equidistant from (2, 5, 5) and (−6, 3, 1) has normal
n = h2, 5, 5i−h−6, 3, 1i = h8, 2, 4i = 2h4, 1, 2i, and passes through the midpoint 12 (h2, 5, 5i+
h−6, 3, 1i) = h−2, 4, 3i. So its equation is
0 = hx + 2, y − 4, z − 3i · h4, 1, 2i = 4(x + 2) + y − 4 + 2(z − 3) − − > 4x + y + 2z = 2.
Sept. 20, 2013
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