Triangle law of vector addition In the triangle law of vector addition, how can we prove that the resultant vector of a and b points in the direction c?

 A: If I understand your question correctly, this is a matter of definition. If you are given two vectors $\vec{AB}$ and $\vec{BC}$, then, by definition, then sub of the two vectors is $\vec{AC}$.
If this is not satisfying, say that you have three points $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ and $C = (c_1, c_2, c_3)$. Then, by definition, the vector from $A$ to $B$ is
$$
\vec{AB} = \pmatrix{b_1 - a_1 \\ b_2 - a_2 \\ b3 - a_3}.
$$
If you use this definition, then it is not hard to show that $\vec{AB} + \vec{BC} = \vec{AC}$.
A: Vector $a$ can be thought of as going from point $A$ to $B$. Similarly, vector $b$ is going from $B$ to $C$ and vector $c$ is going from $A$ to $C$. 
You went from $A$ to $B$ and then from there to $C$. Totally, you went from $A$ to $C$.
So, $a$+$b$=$c$
Note- It's not just in the direction of $c$, but is infact equal to $c$
A: Note we can prove it using physics . let the distance travelled by a particle be $a+b$ according to your figure. But the displacement is $c$ . now the triangle be $90$ at $B$ so we know magnitude if resultant is $\sqrt{a^2+b^2+2abcos(90)}=\sqrt{a^2+b^2}$ also magnitude of $c$ is $√c^2$ so if we square both we get $a^2+b^2,c^2$ but they are equal according to pythagoras theorem therefore not only resultant vector but magnitude of both vectors is the same. Also alternate way is complete the parallelogram and by parallelogram law of vectors the resultant of adjacent vectors  is diagonal . now as |d| which is parallel to $|b|$ are equal hence the result follows .
