Vectors that form a triangle!

Question

I have a problem here.

How can I prove that sum of vectors that form a triangle is equal to 0 $(\vec {AB}+\vec {BC}+\vec {CA}=\vec 0)$ ?

Answer 1

from the triangle law : $\vec {AB}+\vec{BC}=\vec{AC}$

$\vec {AC}$ will be resultant vector of addition of other two vectors.

$\vec {AB}+\vec{BC}=\vec{AC}$

$\vec {AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}$ (add $\vec{CA}$ on both side)

$\vec {AB}+\vec{BC}+\vec{CA}=\vec{AC}-\vec{AC}$

(because when AC and CA are same in magnitude but they are opposite in direction)

$\vec {AB}+\vec{BC}+\vec{CA}=\vec0$

hence proved

Answer 2

well, define $A := (a_1,a_2, \dots ,a_n)$, then $B := (b_1,b_2, \dots ,b_n)$ and $C := (c_1,c_2, \dots ,c_n)$.

Then $\overrightarrow{AB}=(b_1-a_1,b_2-a_2, \dots ,b_n-a_n)$ similarly $\overrightarrow{AC}=(c_1-a_1,c_2-a_2, \dots ,c_n-a_n)$ and $\overrightarrow{BC}=(b_1-c_1,b_2-c_2, \dots ,b_n-c_n)$.

let's make $\overrightarrow{AB}+\overrightarrow{BC}= (b_1-a_1 + c_1-b_1,b_2-a_2+c_2-b_2, \dots ,b_n-a_n+c_n-a_n) = \overrightarrow{AC}=-\overrightarrow{CA}$ and hence the result.

NB if you don't know how to work with vectors, then you may want to take a look here Wiki - vectors. I've worked in $\mathbb{R}^n$ with and affine structure on it.

Answer 3

Vector addition is defined as follows: $$\vec{PQ}+\vec{QR}:=\vec{PR}$$ So that, $\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AA}={\bf 0}$.