I have a vector with defined magnitude and direction and want to project it into two vectors. The resultant of those two new vectors should be equal to the first vectors. All of the 3 vectors have the same beginning point (The zero of the unit circle). The angle of the target vectors is 70 and 110 degrees. How I should do this?
I have done this with vectors of 45 and 135 degrees easily here, but this practice only works with those specific angles.
If this is not possible with projection, then what other ways you recommend?
Note that this operation is going to be executed by a computer at pretty high speed and shouldn't take much time, and the only thing needed is the magnitude of those two vectors. Sorry for my unprofessional language; I barely have a high-school degree right now.
Orthogonal projection will work if the two vectors you project onto are perpendicular to each other. Vectors at $45$ degrees and $135$ degrees work because they are $90$ degrees apart, so they are perpendicular. Vectors at $70$ and 110 degrees are only $40$ degrees apart.
But if what you really want is a vector at $70$ degrees and another at $110$ degrees (or whatever angles happen to be required at a particular time), whose vector sum (or resultant) should equal a given vector, this is a solvable problem.
Suppose you know the direction and length of the vector marked $w$ in the figure below. You also know the directions in which the vectors marked $u$ and $v$ should point, but you do not know the lengths of those vectors.
As you may know, in order for the resultant of $u$ and $v$ to be $w,$ there must be a triangle like one of the two triangles in the figure. Let's use the triangle on the right. Since you know the directions of all three vectors, you can find the angle $\alpha$ between $u$ and $w$ and the angle $\beta$ between $v$ and $w$. The two vectors/segments marked $u$ are parallel, so you also have an angle $\alpha$ as shown at the top of the right-hand triangle. Finally, you can find the angle $\theta$ by knowing that the sum of angles in a triangle is always $180$ degrees.
For example, if the two unknown vectors are at $70$ degrees and $110$ degrees and the known vector is at $85$ degrees, then $\alpha = 110\text{ degrees} - 85\text{ degrees} = 25\text{ degrees},$ $\beta = 85\text{ degrees} - 70\text{ degrees} = 15\text{ degrees},$ and $\theta = 180\text{ degrees} - \alpha - \beta = 140\text{ degrees}.$
(The vectors in the drawing are not at $70$ and $110$ degrees, of course. I drew them in directions that make more room to put the letters.)
Now we bring in a little trigonometry, namely, the Law of Sines. Applied to the right-hand triangle, the Law of Sines gives us a relationship among the three angles $\alpha,$ $\beta,$ and $\theta$ of the triangle and the three sides of length $\lVert u\rVert,$ $\lVert v\rVert,$ and $\lVert w\rVert.$
$$ \frac{\lVert v\rVert}{\sin(\alpha)} = \frac{\lVert u\rVert}{\sin(\beta)} = \frac{\lVert w\rVert}{\sin(\theta)}. $$
So you compute the sine of each angle, and then you know everything in that formula except $\lVert u\rVert$ and $\lVert v\rVert.$ But a little algebra tells us that $$ \lVert u\rVert = \frac{\lVert w\rVert \sin(\beta)}{\sin(\theta)} \qquad\text{and}\qquad \lVert v\rVert = \frac{\lVert w\rVert \sin(\alpha)}{\sin(\theta)}. $$
And those are the lengths you need.
The formulas work for any angles as long as the vector $w$ is "inside" the smaller angle between the two other vectors as shown in the figure. If $w$ is "outside" the angle then you will have to reverse one or both of the other vectors, that is, change their directions by $180$ degrees.
As a reminder, if you are doing this in a computer program or mathematical software then the sine function usually requires its input to be an angle measured in radians, so if you have your angles in degrees you will have to convert them to radians before calling $\sin().$