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

AI Thread Summary
To find a unit vector that bisects the angle between the vectors 5.00i + 11.0j and 2.00i − 1.00j, it is essential to first determine the unit vectors for each original vector. The correct approach involves calculating the unit vectors, then averaging them to find the bisector. The calculations provided in the discussion reveal confusion in combining the unit vectors and deriving the final result. Attention to detail in the math, particularly in the y-component, is crucial for achieving the correct bisector. The solution hinges on accurately applying vector addition and normalization to arrive at the desired unit vector.
mubeen916
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Here's the problem I've been having difficulty with:

Find a vector with a 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.

Someone else had posted this same question before, but the answer that was given was incorrect.( (0.94 i + 0.34 j))

Please help by explaining the solution and how you arrived at it. I appreciate the help!
 
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mubeen916 said:
Please help by explaining the solution and how you arrived at it. I appreciate the help!

Uh-Uh. You have to show what you've tried first.
 
Okay, so I got the sum of both of the vectors to be (7.00i, 10.00j). I took the square root of 7^2+ 10^2, which = 12.2

Now, to get the resulting vector, I used (7.00/12.2i, 10.0/12.2j) to get (.57i, .82j). Where did I go wrong?
 
But you're not looking for the resultant of adding the vectors; You're looking for a vector that is (in terms of angle) halfway between them. Drawing a diagram with labeled angles might help.
 
Actually, I realized I had to find the corresponding unit vectors given for the vectors originally stated in the problem.
To find the unit vectors, I took the first vector (5.00i, 11.0j) and did the following:

5/ square root of 5^2+11^2i, and 11/square root of 5^2 +11^2j, which gave me the unit vector (.414j, .906j).
I did the same for the second vector (2.00i-1.00j).
2.00/square root of 2^2+-1^2i, -1/square root of 2^2+ -1^2j, which gave me the unit vector (.89i, -.447j).

Then I repeated the steps that I wrote out in my previous post using the unit vectors.

.414 + .89/ square root of (.414+.84)^2 +(.906-.447)^2 i, .906+.89/ square root of (.414+.84)^2 +(.906-.447)^2 j

which gave me the final unit vector of (.94i 1.30j). Apparently this is incorrect but I don't understand what I did wrong.
 
Okay, it just 'clicked' as to how your approach to the problem will work. The resultant of two unit vectors will indeed bisect them. That's fine! The difficulty must be with the implementation (i.e. finger problems with the math!).

The value you get for the x-component for the resultant looks okay. Take a closer look at the y-component. In fact, check carefully where the two values you're using in the numerator are coming from...
 
I'm stumped. Physics is not my thing- the answer is probably obvious, but the lightbulb hasn't gone off in my head!
 
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