# Triangle Law of Vector Addition

I know this is a trivial question, but I just wanted to clear my doubt. Using the triangle of vector addition, the resultant of two forces $$F_1$$ and $$F_2$$ is given by $$\sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\phi}$$ In this formula, am I supposed to take only the magnitudes of $$F_1$$ and $$F_2$$ or am I supposed to include their signs as well?

• What do you mean by their sign? They are vectors, they don't have a sign unless you define a sign convention. Also if you are talking about magnitude, then magnitude of a vector is always $>0$.
– user258881
Apr 1, 2020 at 16:03
• I mean, suppose, F1= 5N and F2= −10N, so am I supposed to write √((5²)+(−10²)+2.(5).(-10).cosθ) ?
– user236322
Apr 1, 2020 at 16:10
• You are applying the law of cosines to the sides of a triangle. The magnitude of the forces are all positive.The angle is measured between the two vectors. By the way, the term with the cosine should be negative. Only the components of a vector can be negative, and that's after you define a coordinate system. Apr 1, 2020 at 18:41

See, forces do not have a sign, and for that matter, vectors have no sign. They just have a direction and a magnitude. Their direction is defined using unit vectors and these unit vectors can be arbitrarily chosen. Also the $$\phi$$ in your equation is the angle between two forces. And the $$F_1$$ and $$F_2$$ are the magnitudes of the corresponding forces and magnitudes of vectors are always positive.
We only have to take the magnitude of the forces without sign. But you have to be careful in finding $$\cos \phi$$. In the example of $$F_1= 5 \: \mathrm N$$ and $$F_2= -10 \: \mathrm N$$, the value of $$\phi$$ is $$180^{\circ}$$ or $$\pi$$ and hence $$\cos\phi= -1$$. This means that although you will substitute $$F_1$$ and $$F_2$$ without sign, still the $$2F_1F_2\cos\phi$$ term will be negative.