Problem 2.59 The position vector r goes from point A to a point on the straight line between B and C. Its magnitude is [r]= 6 ft. Express r in terms of scalar components. A (3,5)
Problem 2.59 The position vector r goes from point A to a point on the straight line between B and C. Its magnitude is [r]= 6 ft. Express r in terms of scalar components. A (3,5)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Problem 2.59 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is [r]= 6 ft. Express r in terms of scalar
components.
Strategy
² = AD = AC
CD = d ecB
=
: a) Find vector AD at some
point
b) Find d such that |AD| = 6 ft.
AC + CD
d
+
A (3,5) ft
arbitrary
on CB at a distance & from C.
(7-12)²
B (7,9) ft
5² +62
m
(9-3)²
(7-12)2 + (9-3) 5 = d (√5 [+])
dcas
=d
J
25d²
61
C(12, 3) ft
AC = (12-3) 2² + (8-5)j
92-23
=
5d
:. F² = AD = ( 9 - 51 ) 2 + (-2+61) =
-
| ² | = |AB| = √√(9-58)2²+(-2+64)²06 = 6 *
90d
81-
√61
X
+4- +
24d 36d²
161
61
114
√ET d +49 = 0→d=
+114
167
=
36
- ± √(14² - 4(49)
61
Z
d²_
d = 9,36 or 5₁23 = F² = 5.65 5 + 2.02 J
in
beyond point B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf815a4f-1aaa-489f-9ff2-9168b381a40c%2F5e927b45-641a-4cb5-a98b-7fd9e03ae594%2F408sbks_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 2.59 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is [r]= 6 ft. Express r in terms of scalar
components.
Strategy
² = AD = AC
CD = d ecB
=
: a) Find vector AD at some
point
b) Find d such that |AD| = 6 ft.
AC + CD
d
+
A (3,5) ft
arbitrary
on CB at a distance & from C.
(7-12)²
B (7,9) ft
5² +62
m
(9-3)²
(7-12)2 + (9-3) 5 = d (√5 [+])
dcas
=d
J
25d²
61
C(12, 3) ft
AC = (12-3) 2² + (8-5)j
92-23
=
5d
:. F² = AD = ( 9 - 51 ) 2 + (-2+61) =
-
| ² | = |AB| = √√(9-58)2²+(-2+64)²06 = 6 *
90d
81-
√61
X
+4- +
24d 36d²
161
61
114
√ET d +49 = 0→d=
+114
167
=
36
- ± √(14² - 4(49)
61
Z
d²_
d = 9,36 or 5₁23 = F² = 5.65 5 + 2.02 J
in
beyond point B
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