# Some question about vector

I posted similar question before. But I am yet to be clarified. Suppose I know a vector F which is defined as (xr - x) ,(yr -y), (zr -z). This vector is with angle theta with another vector F1{ (x1 -x), (y1 -y), (z1-z) }. This has been shown in the following figure. Now I have aonther vector F2 defined by ((x2 -x), (y2 -y), z2-z)). which is perpendicular to F1. I need to find (x2, y2, z2). I did as follows:

``````
F2 (x2, y2, z2)
|         *
|       * F (xr, yr, zr)
|    * theta is angle between F and F1
|  *
|*_______  F1(x1, y1, z1)
(x, y,z)

``````

I first normalized F. So F/|F|,
Then I multipled this unit normal by sin(theta).

Now I tried to find the magnitude and I did like this |F|sin(theta).

Now (x2 , y2, z2) for vector F2 is:

x2 = x + ( Fx/|F|)sin(theta) |F|sin(theta);
y2 = y + ( Fy/|F|)sin(theta) |F|sin(theta);
z2 = z + ( Fz/|F|)sin(theta) |F|sin(theta);

Am I right? Could any one tell me? Because I am not getting the correct answer.
Thank you all in advance.

I believe that with the information you provided, it’s not possible to determine x2,y2, and z2. Are they all on the same plane? If not, your assumptions about the angles are incorrect. And I assume you’re trying to find a possible x2,y2 and z2 after being normalized since there’s an infinite number of points on that line that can be normalized to a unit vector.

Now I have another vector F2 defined by ((x2 -x), (y2 -y), z2-z)). which is perpendicular to F1.

ok, you want F2 perpendicular to F1, but what is the role of F here? Perhaps you may have missed mentioning additional requirements you have on F2? As it stands there is an infinite number of vectors that are perpendicular to F1 (every vector in the plane that has F1 as normal vector).

Just for your kind information, I want to say that F2 is at (90 -theta) angle with F2. I forgot to provide this information about F2. It seems that F is the resultant of F1 and F2.

This means that F1, F2 and F have to lie in the same plane, but the length of F2 is still unspecified.

One solution to your problem is to project the endpoint of F onto the plane orthogonal to F1 and use this as endpoint of F2.

I use the abbreviations P for (x,y,z), P1 for (x1,y1,z1), P2 for (x2,y2,z2) and Pr for (xr,xy,zr).

Then F is Pr - P, F1 is P1 - P and F2 is P2 - P.

The plane orthognonal to F1 is the set of all points X which satisfy dot(X-P, F1) = 0. The line parallel to F1 through Pr is given by X = Pr + x * F1 for all x.

The intersection point has to be in both sets, so
0 = dot(Pr + x * F1 - P, F1)
= dot((Pr - P) + x * F1, F1)
= dot(F + x * F1, F1)
= dot(F, F1) + x * dot(F1, F1)

So x is -dot(F, F1) / dot(F1, F1) and by substituting x in the line equation we get P2 = Pr - dot(F, F1) / dot(F1, F1) * F1.

Thank you very much for taking time in explaining the thing.