Vectors A and B lie in an xy plane. A has magnitude 8.00 and angle 130 degree ;
ID: 1328151 • Letter: V
Question
Vectors A and B lie in an xy plane. A has magnitude 8.00 and angle 130 degree ; B has components Bx = -7.72 and By = -9.20. (a) What is 5A -B? What is 4A times 3B in unit-vector notation and magnitude-angle notation with spherical coordinates (see Fig. 3-34)? (d) What is the angle between the directions of A and 4A times 3B? A has the magnitude 12.0 m and is angled 60.0 degree counterclockwise from the positive direction of the .v axis of an xy coordinate system. Also. 5 = (12.0 m)i + (8.00 m)j on that same coordinate system. We now rotate the system counterclockwise about the origin by 20.0 degree to form an x'y' system. On this new system, what are (a) A and (b) B. both in unit-vector notation?Explanation / Answer
magnitude of A = 8.00
angle of A = 130 degree
vector A = 8.00 cos 130 i + 8.00 sin 130 j = -5.14i + 6.128j
vector B = -7.72i -9.20j
(a) 5A.B = 5( -5.14i + 6.128j).( -7.72i -9.20j) = 5(39.68 - 56.38) = -83.488
(b) 4A x 3B = 4( -5.14i + 6.128j)x 3( -7.72i -9.20j) = (-20.56 i + 24.5j)x (-23.16i - 27.6j) = 567.456k + 567.42k = 1134.876 k
(c) In spherical coordinate form 4A x 3B = (r, theta, phi)
r = (x^2 +y^2 +z^2)^1/2 = (0^2 + 0^2 + 1134.876^2)^1/2 = 1134.876
theta = arctan(y/x) = arctan(0/0) = not defined
phi = arccos(z/r) = arccos(1134.876/1134.876) = 0 degrees
4A x 3 B = (1134.876, not defined, 0)
(d) Cross product of two vectors is perpendicular to the plane containing two vectors. 4A x 3 B is perpendicular to both A and B. Hence angle between 4 A x 3B and A is 90 degrees
(e) A + 3.00k = -5.14i + 6.128j + 3.00k
(f) A + 3.00 k = (r, theta, phi)
r = (x^2 +y^2 +z^2)^1/2 = (-5.14^2 + 6.128^2 + 3.00^2)^1/2 = 8.54
theta = arctan(y/x) = arctan(6.128/(-5.14)) = -50 degrees
phi = arccos(z/r) = arccos(3.00/8.54) = 69.4 degrees
In spherical coordinates notation A + 3.00 k = (8.54, -50o, 69.4o)
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