MA 22S1
Assignment 4
Due 16-18 November 2015
Id:
22S1-f2015-4.m4,v 1.3 2016/05/14 00:05:08 john Exp john
1. Find
(a) the angle between the intersecting planes x + y + z + 1 = 0 and
x + 2y + 3z + 4 = 0,
Solution:
cos θ =
|n1 · n2 |
|(1, 1, 1) · (1, 2, 3)
6
=
= √ = 0.92582.
kn1 kkn2 k
k(1, 1, 1)kk(1, 2, 3)k
42
θ = 0.38760 = 22.208◦.
(b) the distance between the parallel planes x + 2y + 3z + 4 = 0 and
x + 2y + 3z − 4 = 0,
Solution: To use the formula from lecture you need points r1
and r2 on the two planes. There are many possible choices, but
(−4, 0, 0) and (4, 0, 0) are fairly simple.
|n · (r1 − r2 )|
(1, 2, 3) · ((−4, 0, 0) − (4, 0, 0))
=
knk
k(1, 2, 3)k
8
= √ = 2.13809.
14
D=
(c) the point of intersection and the angle between the line x = t,
y = t, z = t and the line x = s + 2, y = 2s + 3, z = 3s + 4,
Solution:
To find the point of intersection we solve the six equations above
for x, y, z, s and t. This gives
s = −1,
t = 1,
1
x = 1,
y = 1,
z=1
Id:
22S1-f2015-4.m4,v 1.3 2016/05/14 00:05:08 john Exp john 2
so the intersection is (1, 1, 1). For the angle,
cos θ =
|v1 · v2 |
|(1, 1, 1) · (1, 2, 3)
6
=
= √ = 0.92582.
kv1 kkv2 k
k(1, 1, 1)kk(1, 2, 3)k
42
θ = 0.38760 = 22.208◦.
(d) the distance between the skew lines x = t, y = t, z = t and
x = s + 2, y = 2s + 3, z = 3s − 4,
Solution:
D=
8
|(r1 − r2 ) · (v1 × v2 )|
= √ = 3.26599.
kv1 × v2 k
6
(e) the distance between the parallel lines x = t, y = t, z = t + 1 and
x = s, y = s, z = s − 1.
Solution:
√
5
k(r1 − r2 ) × vk
D=
= √ = 1.29099.
kvk
3
2. Each of the following surfaces is given in one of Cartesian, cylindrical
or spherical coordinates. Convert it to the other two. Simplify as much
as possible.
(a) x2 − y 2 + z 2 = 0,
Solution: In cylindrical coordinates
r 2 cos2 θ − r 2 sin2 θ + z 2 = 0
or
r 2 cos(2θ) + z 2 = 0.
In spherical coordinates
ρ2 sin2 φ cos(2θ) + ρ2 cos2 φ = 0
or
cos(2θ) + cot2 φ = 0.
(b) ρ = cot φ csc φ,
Solution: Multiplying by ρ sin2 φ,
ρ cos φ = ρ2 sin2 φ
Id:
22S1-f2015-4.m4,v 1.3 2016/05/14 00:05:08 john Exp john 3
or, in cylindrical coordinates,
z = r2 .
In Cartesian coordinates
z = x2 + y 2.
(c) r 2 − 2rz cos θ + z 2 = 1,
Solution: In Cartesian coordinates
x2 + y 2 + z 2 − 2xz = 1.
In spherical coordinates
ρ2 − 2ρ2 cos θ sin φ cos φ = 1
or
ρ2 (1 − cos θ sin(2φ)) = 1.