Not quite.
v_{rel}=√(v_{1}^{2}+v_{2}^{2}-2v_{1}v_{2}cos45)
This is the magnitude of the vector with the origin in the head of v_{1} and the head in the head of v_{2}. it's a sliding vector. It's the speed you see the other one moving if you are in one of them. So if you're moving at speed v_{1} with respect to the origin you see the other one moving at v_{1}-v_{2} and the vector's origin must be in the other one's position.
I prefer to use the term "velocity" to mean the vector, "speed" to mean the magnitude of that vector.
"What if you are going 45°North of East at 15 mph and you see another car is going due East at 8mph. What is your velocity compared to the other car?"
The basic idea is to treat the other car as if it were standing still. If you are going 45 degrees North of East at 15 mph, then you "components" (setting up a coordinates system with x to the east and y to the north) are (15 cos(45),15 sin(45))= (15[sqrt](2)/2, 15[sqrt](2)/2). The other car is going due east a 8 mph so it components are (8, 0). "Compare" the two by subtracting the two vectors:
(15[sqrt](2)/2,15[sqrt](2)/2)- (8,0)=(15[sqrt](2)/2-8,15[sqrt](2)/2)
Since [sqrt](2)/2 is about .7, 15[sqrt](2)/2 is about 10.5 and 15[sqrt](2)/2- 8 is about 2.5. The occupants of the second car would see you going mostly east with a little bit of north component.
Since the problem asked for the vector relative to the second car, there is no reason to calculate it "length" or speed. The answer is
(15[sqrt](2)/2-8,15[sqrt](2)/2).
If you are required to give the vector in the same form as the two velocities were originally given, speed and direction, you could use the Pythagorean theorem to find the speed and
tan(theta)=(15[sqrt](2)/2-8)/(15[sqrt](2)/2) to find the direction.
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