# Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors

## Theorem

Let $\mathbf r$ be a vector quantity embedded in space.

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be non-coplanar.

Then $\mathbf r$ can be expressed uniquely as the resultant of $3$ vector quantities which are each parallel to one of $\mathbf a$, $\mathbf b$ and $\mathbf c$.

## Proof

Let $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ be unit vectors in the directions of $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.

Let $O$ be a point in space.

Take $\vec {OP} := \mathbf r$.

With $OP$ as its space diagonal, construct a parallelepiped with edges $OA$, $OB$ and $OC$ parallel to $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ respectively.

Only one such parallelepiped can be so constructed.

Let $x$, $y$ and $z$ be the length of the edges $OA$, $OB$ and $OC$ respectively.

Then:

 $\ds \mathbf r$ $=$ $\ds \vec {OA} + \vec {AF} + \vec {FP}$ $\ds$ $=$ $\ds \vec {OA} + \vec {OB} + \vec {OC}$ $\ds$ $=$ $\ds x \mathbf {\hat a} + y \mathbf {\hat b} + z \mathbf {\hat c}$

Thus $\mathbf r$ is the resultant of the $3$ components $x \mathbf {\hat a}$, $y \mathbf {\hat b}$ and $z \mathbf {\hat c}$ which, by construction, are parallel to $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.

The fact that only one parallelepiped can be constructed in the above proves uniqueness.

$\blacksquare$