Equivalence of Definitions of Vector Sum

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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 $\mathbf u$.

Vector-sum-triangle-law.png

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:


Vector-sum-equivalence.png


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.

$\blacksquare$