Equivalence of Definitions of Vector Sum

Theorem

The following definitions of the concept of Vector Sum are equivalent:

Component Definition

Let $\mathbf u$ and $\mathbf v$ be represented by their components considered to be embedded in a real $n$-space:

 $\ds \mathbf u$ $=$ $\ds \tuple {u_1, u_2, \ldots, u_n}$ $\ds \mathbf v$ $=$ $\ds \tuple {v_1, v_2, \ldots, v_n}$

Then the (vector) sum of $\mathbf u$ and $\mathbf v$ is defined as:

$\mathbf u + \mathbf v := \tuple {u_1 + v_1, u_2 + v_2, \ldots, u_n + v_n}$

Note that the $+$ on the right hand side is conventional addition of numbers, while the $+$ on the left hand side takes on a different meaning.

The distinction is implied by which operands are involved.

Triangle Law

Let $\mathbf u$ and $\mathbf v$ be represented by arrows embedded in the plane such that:

$\mathbf u$ is represented by $\vec {AB}$
$\mathbf v$ is represented by $\vec {BC}$

that is, so that the initial point of $\mathbf v$ is identified with the terminal point of $\mathbf u$.

Then their (vector) sum $\mathbf u + \mathbf v$ is represented by the arrow $\vec {AC}$.

Proof

Let the Cartesian coordinate system be selected such that $\mathbf u$ and $\mathbf v$ are embedded in the $x$-$y$ plane.

Then:

 $\ds \mathbf u$ $=$ $\ds \tuple {u_1, u_2}$ $\ds \mathbf v$ $=$ $\ds \tuple {v_1, v_2}$

and all other components, if there are any, are then zero.

Thus by the component definition:

$\mathbf u + \mathbf v = \tuple {u_1 + v_1, u_2 + v_2}$

$\Box$

Now we use the triangle law:

Let:

 $\ds A$ $=$ $\ds \tuple {A_x, A_y}$ $\ds B$ $=$ $\ds \tuple {B_x, B_y}$ $\ds C$ $=$ $\ds \tuple {C_x, C_y}$

Thus we have:

 $\ds \mathbf u + \mathbf v$ $=$ $\ds \vec {AC}$ $\ds$ $=$ $\ds \tuple {C_x - A_x, C_y - A_y}$ $\ds$ $=$ $\ds \tuple {\paren {C_x - B_x} + \paren {B_x - A_x}, \paren {C_y - B_y} + \paren {B_y - A_y} }$ $\ds$ $=$ $\ds \tuple {v_1 + u_1, v_2 + u_2}$

It is seen that $\mathbf u + \mathbf v$ calculated using the triangle law is the same as $\mathbf u + \mathbf v$ calculated using the component definition.

$\blacksquare$