Math 2433 Homework #8
Solutions
Important: I will only provide helpful hints and not detailed solutions, but
you have to write all the steps involved along with careful explanations.
8, 10, 12, 16, 20, 26, 28, 32
Section 13.1.
Find the lengths of the sides of the triangle P QR. Is it a right-angled triangle? Is it an
isosceles triangle?
8. P (2, −1, 0), Q(4, 1, 1), R(4, −5, 4).
Solution: We have
p
|P Q| = (4 − 2)2 + (1 + 1)2 + 12 = 3
p
√
|QR| = (4 − 4)2 + (−5 − 1)2 + (4 − 1)2 = 45
p
|RP | = (2 − 4)2 + (−1 + 5)2 + (0 − 4)2 = 6
Since |QR|2 = |P Q|2 + |RP |2 , this is a right-angled triangle.
10. Find the distance from (3, 7, −5) to each of the following.
(a) The xy plane, (d) the x axis.
Solution: I will only do two parts here. You have to do all of them. For (a), the
projection of (3, 7, −5) ontopthe x − y plane is (3, 7, 0) and thus the distance from the
point to the x − y plane is (3 − 3)2 + (7 − 7)2 + (−5 − 0)2 = 5. Similarly, the distance
between √
the given point √to the x axis is the distance between (3, 7, −5) and (3, 0, 0),
which is 0 + 49 + 25 = 74.
12. Find an equation of the sphere with center (2, −6, 4) and radius 5, Describe the
intersection with each of the coordinate planes.
Solution: An equation of the given sphere is
(x − 2)2 + (y + 6)2 + (z − 4)2 = 25.
Intersection with the x − y plane is given by
(x − 2)2 + (y + 6)2 + (0 − 4)2 = 25,
2
or
2
(x − 2) + (y + 6) = 9
which is a circle in the x − y plane centered at (2, −6, 0) with radius 3. You can find
the intersection with other coordinate planes in the same manner.
1
2
Solutions
16. Show that the equation represents a sphere, and find its center and radius.
x2 + y 2 + z 2 + 8x − 6y + 2z + 17 = 0.
Solution: Completing the squares, we get
(x2 + 8x + 16) − 16 + (y 2 − 6y + 9) − 9 + (z 2 + 2z + 1) − 1 + 17 = 0
or
(x + 4)2 + (y − 3)2 + (z + 1)2 = 9
which is a sphere with radius 3 and center (−4, 3, −1).
20. Find an equation of the sphere if one of its diameters has endpoints (2, 1, 4) and
(4, 3, 10).
Solution: In order to find an equation of a sphere, we need the radius and the center.
The radius is half the length of the diameter, so
r=
√
1p
(4 − 2)2 + (3 − 1)2 + (10 − 4)2 = 11.
2
The center (h, k, l) is the midpoint of the diameter, so
h=2+
3−1
10 − 4
(4 − 2)
= 3, k = 1 +
= 2, k = 4 +
= 7.
2
2
2
That is, the center is (3, 2, 7). An equation of the given sphere is
(x − 3)2 + (y − 2)2 + (z − 7)2 = 11.
Describe in words the region of R3 represented by the equation or inequality.
26. y ≥ 0.
Solution: It is the region containing all points on and to the right of the x − z plane.
28. z 2 = 1.
Solution: The equation is equivalent to z 2 − 1 = 0 or (z − 1)(z + 1) = 0, which represent
planes parallel to the x − y plane intersecting the z axis at (0, 0, 1) and (0, 0, −1).
32. x2 + y 2 + z 2 > 2z.
Solution: The inequality is x2 + y 2 + z 2 − 2z > 0. By completing the squares, we get
x2 + y 2 + (z − 1)2 > 1.
This inequality represents the region consisting of all points lying outside a sphere of
radius 1 centered at (0, 0, 1).