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19.
lies in the xy-plane, and is
perpendicular to the xy-plane (so is
perpendicular to ). So,
and OB ! OP ! , OA ! ,
form a right triangle and, by
the Pythagorean theorem,
OP ! 2 OA ! 2
Similarly, OA ! OB !
a ! 2
b !
by the
triangle law of vector addition, where
a ! 15, 0, 02 and b ! 10, 10, 02, and
these three vectors form a right
triangle as well. So,
OA ! 2 0a ! 0
2 b ! 2
25 100
125
Obviously OB ! 2 100, and so
substituting gives
OP ! 2 OA ! 2 OB ! 2
125 100
225
Section 6.6, pp. 324–326
1. a.
2.
OP ! 225
15
16, 5, 22
AB ! 13, 22, BA ! 13, 22
A(–1, 3)
BA
b. OA ! ,
c. AB ! 3.16 OB !
3.61,
BA ! 5.39
3.61
30
25
20
15
10
y
y
AB
B(2, 5)
A(6, 10)
5
O(0, 0) OA
–9 –6 –3
0
–5
3 6 9
x
x
0 0
0
0
a.
y
20
(12, 20)
15
10
5
O(0, 0) (3, 5)
–12 –9 –6 –3
0
3 6 9 12
(–3, –5) –5
–10
–15
–20
(–12, –20)
1
b.
2 OA! and 1
2OA ! 2 OA! ,
and 2OA !
3. 5
4. a. a 3,
b 5
b. 5.83
5. a. 0a ! ,
b. 0a ! 61
,
0a ! b ! 0b ! 41
b ! 100.02
28.28
6. a.
b.
12, 72
130, 02
c. (1, 11)
7. a. 7i ! 8j !
b. 3i ! 51j !
c. 29i ! 28j !
8. a. about 4.12
b. about 6.71
c. about 18.38
9.
d. about
a. AB ! 18.38
, ,
EF ! 14, 22 CD !
16, 42,
GH ! 12, 42
15, 02
b. AB ! 4.47, CD ! 4.47,
EF ! 7.21,
GH ! 5
10. a. OC ! 117, 32, BA ! 15, 92,
11. a.
BC ! 16, 32
b. OA ! 16, 32 BC ! , so obviously
we will have OA ! BC ! .
C(–4, 11)
b. AB ! 14, 32, AB ! 5,
AC ! 10,
CB ! 11.18
c. CB ! 2 125, AC ! 2 100,
AB ! 2 25
y
B(6, 6)
A(2, 3)
Since CB ! 2 AC ! 2 AB ! 2 , the
triangle is a right triangle.
x
x
12. a.
y
C(2, 8)
A(–1, 2)
x
B(7, –2)
b.
y
X(–6, 12)
C(2, 8)
Z(10, 4)
A(–1, 2)
x
B(7, –2)
Y(4, –8)
c. X16, 122, Y14, 82,
Z110, 42
13. a. x 7,
y 2
b. x 12,
y 4.
14. a.
y
C(x, y)
B(–6, 9)
b. Because ABCD is a rectangle, we
will have
BC ! AD !
1x, y2 16, 92 18, 112 12, 32
1x 6, y 92 16, 82
x 6 6
y 9 8
So, x 0 and y 17, i.e., C10, 172.
15. a. P a 21
10 , 0 b
b. Q a 0, 21 4 b
16. a 9 41 , 40
41 b
17. a. about 80.9°
b. about 35.4°
D(8, 11)
A(2, 3)
x
OA ! OB ! Answers 667
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