Resolution of a Vector in Three Dimensions

Setting up in space. Let $\hat i,\hat j,\hat k$ be unit vectors along the positive $x,y,z$ axes, all with initial point at the origin $O$. Let $P(x,y,z)$ be any point in space.

Theorem. $\vec{OP}$ can be written uniquely as $\vec{OP}=x\hat i+y\hat j+z\hat k,\qquad |\vec{OP}|=\sqrt{x^2+y^2+z^2}.$

Proof (sketch). Let $Q$ be the foot of the perpendicular from $P$ to the $xy$-plane, and let $R,S$ be the feet of the perpendiculars from $Q$ to the $x$- and $y$-axes. Then $OR=x,\ OS=y,\ QP=z$, and $\vec{OP}=\vec{OQ}+\vec{QP}=\vec{OR}+\vec{RQ}+\vec{QP}=x\hat i+y\hat j+z\hat k$. For the magnitude: in right triangle $ORQ$, $OQ^2=OR^2+RQ^2=x^2+y^2$; and in right triangle $OQP$, $OP^2=OQ^2+QP^2=x^2+y^2+z^2$, giving $|\vec{OP}|=\sqrt{x^2+y^2+z^2}$.

Three non-coplanar vectors span space uniquely. If $\vec a,\vec b,\vec c$ are three non-coplanar vectors, then any vector in space can be written as $\lambda\vec a+\mu\vec b+\nu\vec c$ in exactly one way, for some scalars $\lambda,\mu,\nu$. …