# Equality of Vector Quantities

## Theorem

Two vector quantities are equal if and only if they have the same magnitude and direction.

That is:

$\mathbf a = \mathbf b \iff \paren {\size {\mathbf a} = \size {\mathbf b} \land \hat {\mathbf a} = \hat {\mathbf b} }$

where:

$\hat {\mathbf a}$ denotes the unit vector in the direction of $\mathbf a$
$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$.

## Proof

Let $\mathbf a$ and $\mathbf b$ be expressed in component form:

 $\ds \mathbf a$ $=$ $\ds a_1 \mathbf e_1 + a_2 \mathbf e_2 + \cdots + a_n \mathbf e_n$ $\ds \mathbf b$ $=$ $\ds b_1 \mathbf e_1 + b_2 \mathbf e_2 + \cdots + b_n \mathbf e_n$

where $\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n$ denote the unit vectors in the positive directions of the coordinate axes of the Cartesian coordinate space into which $\mathbf a$ has been embedded.

Thus $\mathbf a$ and $\mathbf b$ can be expressed as:

 $\ds \mathbf a$ $=$ $\ds \tuple {a_1, a_2, \ldots, a_n}$ $\ds \mathbf b$ $=$ $\ds \tuple {b_1, b_2, \ldots, b_n}$

We have that:

 $\ds \size {\mathbf a}$ $=$ $\ds \size {\tuple {a_1, a_2, \ldots, a_n} }$ $\ds$ $=$ $\ds \sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} }$

and similarly:

 $\ds \size {\mathbf b}$ $=$ $\ds \size {\tuple {b_1, b_2, \ldots, b_n} }$ $\ds$ $=$ $\ds \sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} }$

Also:

 $\ds \hat {\mathbf a}$ $=$ $\ds \widehat {\tuple {a_1, a_2, \ldots, a_n} }$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} } } \mathbf a$

and similarly:

 $\ds \hat {\mathbf b}$ $=$ $\ds \widehat {\tuple {b_1, b_2, \ldots, b_n} }$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} } }$

Let $\mathbf a = \mathbf b$.

Then by Equality of Ordered Tuples:

$(1): \quad a_1 = b_1, a_2 = b_2, \ldots a_n = b_n$

Then:

 $\ds \size {\mathbf a}$ $=$ $\ds \sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} }$ $\ds$ $=$ $\ds \sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} }$ from $(1)$ $\ds$ $=$ $\ds \size {\mathbf b}$

and:

 $\ds \hat {\mathbf a}$ $=$ $\ds \dfrac 1 {\sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} } } \mathbf a$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} } } \mathbf b$ from $(1)$ $\ds$ $=$ $\ds \hat {\mathbf b}$