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Find a vector with magnitude of 1.00 that bisects the angle between the vectors 5.00i + 11.0j and 2.00i - 1.00j. Give your answer in rectangular coordinates.
I'm not going to lie, I am terrible at physics and I had no idea how to solve this question, so I asked someone for help. I came out with a long sheet of equations I never heard of nor what was mentioned in my textbook and a final answer of (.94, .34). I don't know if that is right. I never heard of a dot product, never heard of a lot of stuff they used. Can anyone verify if this answer looks right...? (and if you have the smallest inclination, if you know a different way, to explain this, that wouldn't be bad either )
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Let vector A = 5.00i + 11.0j and vector B = 2.00i - 1.00j
_____
Magnitude of A = A = √52+112 = 12.08
____
Magnitude of B = B = √22+12 = 2.24
Dot product of the two vectors A = Ax i + Ay j and B = Bx i + By j is given by:
A.B = ABcosθ …….(1) where θ is the angle between the two vectors
and A.B = AxBx +AyBy ………(2)
Let us first find dot product of the given vectors using (1) and (2):
Using (1) we get: A.B = ABcosθ = 12.08 x 2.24 cosθ = 27cosθ ……(3)
Using (2) we get: A.B = 5 x 2 - 11 x 1 = -1 ……..(4)
Equating (3) and (4): 27cosθ = -1 or cosθ = -1/27 = 0.037 or θ = 92O
Bisector vector will divide the angle between the given vectors into two equal angles i.e. the angle between the bisector vector and anyone of the given vectors shall be 46O.
Let the bisector vector be represented by: C = Cx i + Cy j
Further, we know that the bisector vector is a unit vector. Hence, the magnitude of vector C is 1. Hence,
________
√ Cx2 + Cy2 = 1 or Cx2 + Cy2 = 1 ………(5)
Further, we find the dot product between vector A and vector C using (1) and(2):
Using (1) we get: A.C = ACcos(θ/2) = 12.08 x 1 cos46O = 8.39
Using (2) we get: A.C = 5Cx + 11Cy
Equating we get: 5Cx+11Cy = 8.39 …….(6)
From (5) Cy2 = 1 - Cx2 …….(7)
From (6) Cy2 = (8.39 - 5Cx)2/112 = (8.39 - 5Cx)2/121 …….(8)
Equating (7) and (8): (8.39 - 5Cx)2/121 = 1 - Cx2
70.39 + 25Cx2 – 83.9Cx = 121 - 121Cx2
146Cx2 – 83.9Cx – 50.61 = 0
_______________________
Cx = [- (– 83.9) + √(– 83.9)2 – 4(146)( – 50.61)]/2(146)
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Cx = [+83.9 + √7039 + 29556]/292
Cx = [+83.9 + 191.3]/292
Taking + sign: Cx = 0.94
Substituting for Cx in (7): Cy2 = 1 - 0.942 = 0.1164 or Cy = 0.34
Hence, the bisector unit vector is (0.94 i + 0.34 j)
I'm not going to lie, I am terrible at physics and I had no idea how to solve this question, so I asked someone for help. I came out with a long sheet of equations I never heard of nor what was mentioned in my textbook and a final answer of (.94, .34). I don't know if that is right. I never heard of a dot product, never heard of a lot of stuff they used. Can anyone verify if this answer looks right...? (and if you have the smallest inclination, if you know a different way, to explain this, that wouldn't be bad either )
==================================================================
Let vector A = 5.00i + 11.0j and vector B = 2.00i - 1.00j
_____
Magnitude of A = A = √52+112 = 12.08
____
Magnitude of B = B = √22+12 = 2.24
Dot product of the two vectors A = Ax i + Ay j and B = Bx i + By j is given by:
A.B = ABcosθ …….(1) where θ is the angle between the two vectors
and A.B = AxBx +AyBy ………(2)
Let us first find dot product of the given vectors using (1) and (2):
Using (1) we get: A.B = ABcosθ = 12.08 x 2.24 cosθ = 27cosθ ……(3)
Using (2) we get: A.B = 5 x 2 - 11 x 1 = -1 ……..(4)
Equating (3) and (4): 27cosθ = -1 or cosθ = -1/27 = 0.037 or θ = 92O
Bisector vector will divide the angle between the given vectors into two equal angles i.e. the angle between the bisector vector and anyone of the given vectors shall be 46O.
Let the bisector vector be represented by: C = Cx i + Cy j
Further, we know that the bisector vector is a unit vector. Hence, the magnitude of vector C is 1. Hence,
________
√ Cx2 + Cy2 = 1 or Cx2 + Cy2 = 1 ………(5)
Further, we find the dot product between vector A and vector C using (1) and(2):
Using (1) we get: A.C = ACcos(θ/2) = 12.08 x 1 cos46O = 8.39
Using (2) we get: A.C = 5Cx + 11Cy
Equating we get: 5Cx+11Cy = 8.39 …….(6)
From (5) Cy2 = 1 - Cx2 …….(7)
From (6) Cy2 = (8.39 - 5Cx)2/112 = (8.39 - 5Cx)2/121 …….(8)
Equating (7) and (8): (8.39 - 5Cx)2/121 = 1 - Cx2
70.39 + 25Cx2 – 83.9Cx = 121 - 121Cx2
146Cx2 – 83.9Cx – 50.61 = 0
_______________________
Cx = [- (– 83.9) + √(– 83.9)2 – 4(146)( – 50.61)]/2(146)
___________
Cx = [+83.9 + √7039 + 29556]/292
Cx = [+83.9 + 191.3]/292
Taking + sign: Cx = 0.94
Substituting for Cx in (7): Cy2 = 1 - 0.942 = 0.1164 or Cy = 0.34
Hence, the bisector unit vector is (0.94 i + 0.34 j)