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Solutions to Homework from Sections 9.1 - 9.
Math 234, Spring 2008
Section 9.
- The projection of (2, 3 , 5) to the xy-plane is (2, 3 , 0); to the yz-plane is (0, 3 , 5); and to the xz-plane is (2, 0 , 5). Below
is the box determined by the origin and (2, 3 , 5).
H2, 3, 5L
H0,0,0L
H2, 0, 0L
H2, 3, 0L
H0, 3, 0L
H0, 0, 5L
H2, 0, 5 L
H0, 3, 5 L
The length of the diagonal is
2
2
2
- The equation of the sphere with center (2, − 6 , 4) and radius 5 is:
(x − 2)
2
2
2
= 25
The intersection with the xy-plane is when z = 0, so (x − 2)
2
2
2
= 25, or (x − 2)
2
2
= 9.
This is the circle in the xy-plane with center (2, − 6 , 0) and radius 3.
The intersection with the yz-plane is when x = 0, so (0 − 2)
2
2
2 = 25, or (y + 6)
2
2 = 21.
This is the circle in the yz-plane with center (0, − 6 , 4) and radius
The intersection with the xz-plane is when y = 0, so (x − 2)
2
2
2 = 25, or (x − 2)
2
2 = −11.
Since a sum of squares cannot be negative, there are no solutions to this equation, so the sphere does not intersect the
xz-plane.
- Show that set of points P = (x, y, z) whose distance from A(− 1 , 5 , 3) is twice their distance from B(6, 2 , −2) is a sphere.
The distance from P to A is given by
(x + 1)
2
2
2 , and the distance from P to B is given by
√
(x − 6)
2
2
2
. So we get the equation:
(x + 1)
2
2
2 = 2
(x − 6)
2
2
2
=⇒ (x + 1)
2
2
2
= 4((x − 6)
2
2
2
)
=⇒ x
2
2
− 10 y + 25 + z
2
− 6 z + 9 = 4(x
2
− 12 x + 36 + y
2
− 4 y + 4 + z
2
=⇒ x
2
2
− 10 y + z
2
− 6 z + 35 = 4x
2
− 48 x + 4y
2
− 16 y + 4z
2
=⇒ 0 = 3x
2
− 50 x + 3y
2
− 6 y + 3z
2
=⇒ 3(x
2
x + y
2
− 2 y + z
2
z + 47) = 0
=⇒ x
2
−
x + y
2
− 2 y + z
2
z + 47 = 0
Now we complete the squares on the left hand side:
x −
2
2
2
− 1 +
z −
2
2
x −
2
2
z −
2
2
2
We are left with the equation for a sphere: specifically, the sphere with center
25
3
11
3
and radius
Section 9.
0 x
0
1
2
3
4
y
0
1
2
z
0
1
2
3
4
y
- The situation is pictured below:
20 40 60 80 100 120
50
100
150
200
wind
airspeed
The vector for the wind speed is 〈50 cos(
◦
), −50 sin(
◦
)〉 = 〈 25
2 〉 (note that the wind is blowing from the
northwest, not towards the northwest). The vector for the plane’s airspeed is 〈250 cos(
◦
), 250 sin(
◦
)〉 = 〈 125
(the direction of the plane is 60
◦
from due north, so 30
◦
degrees from due east, the x-axis). The resultant of these
vectors 〈 125
To find the true course, we need the angle from due north. We find this using the inverse tangent:
arctan
◦
This is the angle from due east. So the angle from north is 90
◦
− 19. 6
◦
= 70. 4
◦
. So the true course is N70. 4
◦
E.
The ground speed is the magnitude of the resultant:
2
2 = 241.93 km/h.
Section 9.
i − 2
k) × (
j +
k) = (
i ×
j) + (
i ×
k) − 2(
k ×
j) − 2(
k ×
k) =
k + (−
j) − 2(−
i) − 2(
i −
j +
k = 〈 2 , − 1 , 1 〉.
0
1
2
x
0
1 y
0
1
z
a
b
a x b
H0,0,0L
0
1 y
- (a) Consider the points P (2, 1 , 5), Q(− 1 , 3 , 4) and R(3, 0 , 6). The plane containing this points contains the (non-
parallel) vectors
P Q = 〈− 3 , 2 , − 1 〉 and
P R = 〈 1 , − 1 , 1 〉. The cross product of these two vectors is perpendicular
to them both, and therefore to the plane containing them.
〈− 3 , 2 , − 1 〉 × 〈 1 , − 1 , 1 〉 = 〈 2 − 1 , −3 + 1, 3 − 2 〉 = 〈 1 , − 2 , 1 〉
(b) The length of this vector is the area of the parallelogram determined by
P Q and
P R, which is twice the area of
the triangle 4 P QR. So the area of the triangle is
1
2
1
2
√
6
2
- (a) Let P be a point not on the plane containing points Q, R, S. Let ~a =
QR,
b =
QS and ~c =
QP. Show that the
distance d from P to the plane is:
d =
|~a · (
b × ~c)|
‖~a ×
b‖
Recall that |~a · (
b × ~c)| is the volume of the parallelipiped determined by P, Q, R, S, and ‖~a ×
b‖ is the area of the
parallelogram determined by Q, R, S which is the base of the parallelipiped. The distance from P to the plane is
the height of the parallelipiped, which is given by its volume divided by the area of the base.
(b) The distance from the point P (2, 1 , 4) to the plane through the points Q(1, 0 , 0), R(0, 2 , 0) and S(0, 0 , 3) is given
by:
d =
|〈− 1 , 2 , 0 〉 · (〈− 1 , 0 , 3 〉 × 〈 1 , 1 , 4 〉)|
‖〈− 1 , 2 , 0 〉 × 〈− 1 , 0 , 3 〉‖
