How can I find a unit vector that bisects the angle between two other vectors?

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SUMMARY

The discussion focuses on finding a unit vector that bisects the angle between two vectors: A = 5.00i + 11.0j and B = 2.00i - 1.00j. The final answer derived is (0.94, 0.34), confirmed through calculations involving the dot product and the properties of unit vectors. The process includes determining the magnitudes of both vectors, calculating the dot product, and applying algebraic methods to solve for the components of the bisector vector. The use of the dot product is essential for understanding the angle between the vectors and deriving the bisector.

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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 :redface:)
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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)
___________
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)
 
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http://www.falstad.com/dotproduct/

That applet may help you understand the dot product, otherwise, do read about it online. The dot product is useful in this case because it involves the angle between the two vectors. The work that you have there uses the two ways of solving for the dot product [ (1) when you know the magnitudes and the angle between the vectors, (2) when you know the component form] and equates them in order to solve for the angle. Once you have the angle, you can divide by two to find the angle between either vector and the unit vector you're looking for (definition of bisector).

The work from there uses that fact and a few properties of the unit vector: the sum of the squares of the components of the unit vector should be 1 because the magnitude is 1. Then it's using the dot product all over again, noting that you have both the magnitudes of one vector (the work uses vector A, you could have just as easily used vector B) and the unit vector and the angle between them. The rest of the work is just a lot of algebra, and a use of quadratic formula.

The answer looks right to me. It's just a matter of you understanding the dot product and its usefulness in this circumstance. I attached a printscreen of a maple document that I made to illustrate this. The blue line is vector A (5,11), the red line is vector B (2,-1), and the black line is the bisecting unit vector (0.94, 0.34). I hope that assists you in picturing the situation. The dot product was necessary from the beginning in order to determine what angle you're bisecting.

Please ask some specific questions about what in the work confuses you (or the notation) so it may be easier to understand in the future. (e.g. Cx2 and Cy2 in the work are the x- and y- components of C (the unit vector), respectively, squared.)
 

Attachments

The problem statement does not state explicitly that you must use vector operations (dot products, etc.) to solve it, and it looks simple enough to attack with simple geometry.

1. Draw a diagram of the vectors.
2. Find the angles of vectors A and B with respect to the positive x-axis.
3. Determine the angle that bisects them (simple arithmetic average!)
4. Construct a unit vector with that angle. Hint: sin2 + cos2 = 1
 

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