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.
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.
$\blacksquare$
Sources
- 1921: C.E. Weatherburn: Elementary Vector Analysis ... (previous) ... (next): Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $6$. Resolution of a vector
- 1992: Frederick W. Byron, Jr. and Robert W. Fuller: Mathematics of Classical and Quantum Physics ... (previous) ... (next): Volume One: Chapter $1$ Vectors in Classical Physics: $1.2$ The Resolution of a Vector into Components