Find Vector Bisecting Angle Between 3i+4j and 8i-15j

[Andy13]

Homework Statement

Find a vector v that bisects the smaller of the two angles formed by 3i+4j and 8i-15j.

Homework Equations

dot product: cos(theta)= u*v/(||u|| ||v||)

The Attempt at a Solution

I first found the angle in question by putting the two given vectors into the dot product cosine formula. This was the smallest angle.

My next step was to try to do the same thing with some unknown vector, one of the knowns, and the angle I just found, but it turned into an algebraic nightmare. Thoughts? Method is below, except without a true value for theta or cos theta because I forgot what it was off the top of my head-- I've been trying to think about how it would work without numbers first.

Known: angle theta, vector <3,4>
Assume unknown vector (ai+bj)

cos(theta)=((3a+4b)/(5*sqrt(a^2+b^2))

Thanks.

 

[Dick]

If you have two vectors a and b where |a|=|b|, then (a+b)/2 bisects the angle between a and b.

 

[Andy13]

If you have two vectors a and b where |a|=|b|, then (a+b)/2 bisects the angle between a and b.

Thanks Dick. Perhaps I'm missing the point, though-- how would I apply this when my two given vectors are of different magnitudes?

 

[loganblacke]

bump to the top, I need help with a very similar question, and no the magnitudes are not equal.

 

[Dick]

bump to the top, I need help with a very similar question, and no the magnitudes are not equal.

Make them equal. If you divide a vector by its length you get a vector of unit length. And you don't change the direction. Sorry to Andy13 for not getting back earlier. For some reason I never caught the response.

 

[loganblacke]

Thanks!

 

[Dick]

ok so add the unit vectors then divide by two. Then does the resultant vector go from the terminal point of the fist vector to the terminal point of the second vector?

Did you draw a picture? It should go from the origin to the average of the two vectors. The average of the two vectors is the midpoint of the base of the isosceles triangle the two vectors make. That's why it's a bisector.

 

[loganblacke]

Did you draw a picture? It should go from the origin to the average of the two vectors. The average of the two vectors is the midpoint of the base of the isosceles triangle the two vectors make. That's why it's a bisector.

Yeah sorry, as soon as I wrote that I realized my error, I was for whatever reason interpreting "bisect both vectors" as pass through them, not split them.

 

[DikshaBehl]

vectors

I am new to vectors and i am facing some problems in finding the angle between two vectors in multiplications if you can help me in finding it by the help of a formula and a simple example i would be grateful to you...
and one more thing when do we use right hand thumb (screw)rule??

 

[DikshaBehl]

I am new to vectors and i am facing some problems in finding the angle between two vectors in multiplications if you can help me in finding it by the help of a formula and a simple example i would be grateful to you...
and one more thing when do we use right hand thumb (screw)rule??
i know i should have not used your post for posting it but i am not able to post my thered!

 

[loganblacke]

Here's an example for you:

Find a vector W that bisects the smaller of the two angles formed by 12i+5j and 15i - 8j.

Step 1: rewrite as vector A = <12, 5> and Vector B = <15, -8> and find the unit vector for both A and B.

Unit Vector = A/|A| = (12, 5)/Sqrt(12^2+5^2) = (12/13, 5/13)
Unit Vector B = B/|B| = (15, -8)/Sqrt(15^2+-8^2) = (15/17, -8/17)

Step 2: Plug into the formula (Unit Vector A + Unit Vector B)/2 = W

((12/13)+(15/17), (5/13)-(8/17))/2 = W = <399/442, -19/442>