(8c3p28) Given two vectors a = 3.0i-3.9] and b-6.6i +8.2j what is the magnitude
ID: 1661669 • Letter: #
Question
(8c3p28) Given two vectors a = 3.0i-3.9] and b-6.6i +8.2j what is the magnitude (in m) of the vector a? Submit Answer Tries 0/8 Find the direction (in.. deg) of the vector a Submit Answer Tries 0/8 Find the magnitude of the vector b Submit Answer Tries 0/8 Find the direction (in °) of the vector b Submit Answer Tries 0/8 Find the magnitude of the vector a+b Submit Answer Tries 0/8 Find the direction of the vector a+b Submit Answer Tries 0/8 Find the magnitude of the vector b-a Submit Answer Tries 0/8 Find the direction of the vector b-a Submit Answer Tries 0/8 Find the magnitude of the vector a- Submit Answer Tries 0/8 Find the direction of the vector a-b Submit Answer Tries 0/8Explanation / Answer
magnitude of vector a : sqrt((3)2 + (- 3.9)2) = 4.92
direction of vector "a" : 360 - tan-1(3.9/3) = 307.6 degree counterclockwise from +X-axis
magnitude of vector b : sqrt((6.6)2 + (8.2)2) = 10.53
direction of vector "b" : tan-1(8.2/6.6) = 51.12 degree counterclockwise from +X-axis
a + b = 3 i - 3.9 j + 6.6 i + 8.2 j = 9.6 i + 4.3 j
magnitude of vector "a + b" : sqrt((9.6)2 + (4.3)2) = 10.52
direction of vector "a + b" : tan-1(4.3/9.6) = 24.23 degree counterclockwise from +X-axis
b - a = 6.6 i + 8.2 j - 3 i + 3.9 j = 3.6 i + 12.1 j
magnitude of vector "b - a" : sqrt((3.6)2 + (12.1)2) = 12.62
direction of vector "b - a" : tan-1(12.1/3.6) = 73.43 degree counterclockwise from +X-axis
a - b = 3 i - 3.9 j - 6.6 i - 8.2 j = - 3.6 i - 12.1 j
magnitude of vector "a - b" : sqrt((- 3.6)2 + (- 12.1)2) = 12.62
direction of vector "a - b" : 180 + tan-1(12.1/3.6) = 253.42 degree counterclockwise from +X-axis
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