Equality of Vector Quantities

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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 $(1)$
\(\ds \) \(=\) \(\ds \hat {\mathbf b}\)



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