
#1
Mar2109, 09:22 PM

P: 4

Hello,
I have a problem I've been trying to solve. I originally have vector AB. I know both vector A, vector B, and the angle between. They are on the same plane. Now say I change the angle between the two vectors. They still remain the same magnitudes, and remain on the same plane. Assume vector A is fixed. How can I find the new updated vector B. My attempt at a solution: ABcos theta = A*B A, B, and theta are known Equation 1) ABcos theta = axbx + ayby + azbz Equation 2) ax + by + cz = d This is the equation of the plane that the vectors are located on. My problem is I need one more equation to solve this. Maybe I am going about this the wrong way? Any help is appreciated. Thanks, AS 



#2
Mar2209, 12:13 AM

P: 288

Whoa, hold up.
"I originally have vector AB." I'll assume that, by this, you mean you know vector A, and you know vector B, in some sort of coordinate system... say, polar, for convenience. It just so happens that the vectors have an angle between them which can be found by subtracting angle_B  angle_A = angle_difference. Then if the angle is changed so that angle_difference' = angle_difference + angle_delta, then angle_B' = angle_B + angle_delta... no problem. If you have the vectors in cartesian form... well, convert to polar, then convert back when you're done using the above method. If you don't originally know angle_A, then just call angle_A = 0, and go from there. Maybe I'm just missing some subtlety here... 



#3
Mar2209, 12:38 AM

P: 23

Hi,
First, you have to convert your second equation into a an equation with bx, by, bz. For your third equation how about one saying that the length of the new vector B is the same as the length as the original B. Another way you might think about the problem is to find the direction about which you want to rotate B. If you want to keep B coplanar with A then rotate B about the axis perpendicular to both A and B. Then the wikipedia page "Rotation Matrix" has the explicit matrix which will rotate a vector by any angle about an arbitray axis. It's a horrible looking matrix multiplication but you would just be plugging in numbers and doing arithmetic rather than solving a system of equations. 



#4
Mar2209, 04:43 PM

P: 4

For vector AB, how to find vector B given vector A and the angle betweensome scalar = x^2 + x + y^2 + y + z^2 + z. I am unsure how to use this equation with the other 2. 



#5
Mar2309, 02:37 AM

P: 330

I think the solution won't be unique. Imagine the plane to be z=1 and the point A is (0,0,1). Given an angle \theta we want to look for the point (x,y,1) such that [tex]\tan^2 \theta = x^2 + y^2[/tex]. There will be infinite solutions. 



#6
Mar2309, 06:46 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,896

I assume you have some vector A, which, in a given coordinate system, can be represented as [itex]\left(x_0, y_0)\right)[/itex], and want to find vector A, having the same length as A but at angle [itex]\theta[/itex] to A.
Let [itex]\phi[/itex] be the angle A makes with the xaxis. Then, assuming that [itex]\theta[/itex] is measured counterclockwise from A, B makes angle [itex]\theta+ \phi[/itex] with the xaxis. Its components are [itex](Acos(\theta+ \phi),Asin(\theta+ \phi)[/itex]. Now use the facts that [itex]cos(\theta+ \phi)= cos(\theta)\cos(\phi) sin(\theta)sin(\phi)[/itex], [itex]sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)[/itex], [itex]cos(\phi)= x_0/A[/itex], and [itex]sin(\phi)= y_0/A[/itex]. 


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