Chapter_2_examples - Example2.5 Find the angle between A= 2...

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Example 2.1: A man walks 5 m at 37 0 north of east and then 10m at 60 0 west of north. What is the magnitude and direction of his net displacement? R x =A x + B x =5cos37 0 -10sin60 0 = -4.66m R y =A y +B y =5sin37 0 +10cos60 0 = 8.00m R=(R x 2 +R y 2 ) 0.5 = 9.26m tanθ = R y /R x = 8/-4.66= -1.72 Θ could be 120 0 or –60 0 , since R x is negative and R y is positive, the vector lies in the second quadrant and so Θ =120 0 to the +x axis. The net displacement is 9.26m at 30 0 west of north
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Example 2.2: A girl walks 3m east and then 4m south. What is her net displacement? A = 3 i B = -4 j R = A + B = 3 i – 4 j R = (3 2 + 4 2 ) ½, R x is positive, R y is negative tanθ = - 4/3, θ = -53 0 or 143 0
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Example 2.3: A=(2 2 +3 2 +6 2 ) 0.5 =7 m B=(1 2 +2 2 +3 2 ) 0.5 =3.74 m (a) A+B= 10.7 m (b) = (2+1) i + (-3+2) j + (6-3) k =3i -1j+3k B A + B A c 3 2 ) ( ; B A (b) ; B (a)A : find m, k 3 - j 2 i B and m k 6 j 3 - i 2 A vectors Given the - + + + = + = m k j i k j i k j i B A c m B A ˆ 21 ˆ 12 ˆ ) ˆ 3 ˆ 2 ˆ ( 3 ) ˆ 6 ˆ 2 ˆ 2 ( 2 3 2 ) ( 36 . 4 ) 3 1 3 ( 5 . 0 2 2 2 + - = -
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Unformatted text preview: Example2.5 : Find the angle between A= 2 i + j + 2 k and B= 4 i - 3 j • Cosθ= • = AB B A • 5 3 3 1 4 2 x x x--= 1/3 Example2.6: Derive the law of cosines using the scalar product θ cos 2 2 ) ( ) ( 2 2 2 2 2 AB B A C B A B A B A B A C C-+ = ⋅-+ =-⋅-= • Example 2.7: Find the vector product of A = 3i- 2j + k and B = i+4j –2k k j i j i k j k k j i k j i B A ˆ 14 ˆ 7 ) ˆ 4 ˆ ( ) ˆ 4 ˆ 2 ( ) ˆ 6 ˆ 12 ( ) ˆ 2 ˆ 4 ˆ ( ) ˆ ˆ 2 ˆ 3 ( + =-+ + + + =-+ × +-= × Example 2.8: Derive the law of sines using the cross product B A n BC n AC C B C A C C β α sin sin ˆ sin ˆ sin =-= ×-× = ×...
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