Cartesian components and direction cosinesCartesianComponentsAndDirectionCosines2026-08-21 00:35:592026-08-21 00:35:59Definition% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Cartesian Components and Direction Cosines}
Choose a right-handed orthonormal Cartesian basis
\[
\hat{i},\hat{j},\hat{k}.
\]
Every vector has the unique component form
\[
\boxed{\mathbf{u}=u_x\hat{i}+u_y\hat{j}+u_z\hat{k}.}
\tag{1}
\]
\begin{center}
\includegraphics[width=.68\textwidth]{brand_fig_11.png}
\textit{Figure 11, modernized: decomposition into Cartesian components.}
\end{center}
For a point $P=(x,y,z)$,
\[
\mathbf r_P=x\hat{i}+y\hat{j}+z\hat{k}.
\]
Hence
\[
\overrightarrow{P_1P_2}
=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}.
\]
The magnitude is
\[
\lVert\mathbf{u}\rVert^2=u_x^2+u_y^2+u_z^2.
\tag{2}
\]
If $\alpha,\beta,\gamma$ are the direction angles,
\[
u_x=\lVert\mathbf{u}\rVert\cos\alpha,\quad
u_y=\lVert\mathbf{u}\rVert\cos\beta,\quad
u_z=\lVert\mathbf{u}\rVert\cos\gamma,
\]
so
\[
\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.
\]
For planar vectors it is better numerically to use
\[
\theta=\operatorname{atan2}(u_y,u_x)
\]
rather than reconstructing the quadrant from $\tan\theta=u_y/u_x$.
\subsection*{Source examples}
For $\mathbf{u}=(-2,1,2)$,
\[
\lVert\mathbf{u}\rVert=3,
\]
with direction cosines $(-2/3,1/3,2/3)$.
If $\mathbf{u}=(5,2)$ and $\mathbf{v}=(-3,-4)$, then
\[
\mathbf{w}=\mathbf{u}+\mathbf{v}=(2,-2),
\qquad \lVert\mathbf{w}\rVert=\sqrt8,
\]
and $\theta=-45^\circ$ (equivalently $315^\circ$).
A line through $P_1=(x_1,y_1,z_1)$ parallel to
$\boldsymbol\ell=(a,b,c)$ can be written parametrically as
\[
\mathbf r=\mathbf r_1+\lambda\boldsymbol\ell,
\]
or, when the denominators are nonzero,
\[
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.
\]
\section*{Source problems}
\begin{enumerate}
\item Find $\mathbf{u}+\mathbf{v}+\mathbf{w}$ graphically and analytically for:
\begin{enumerate}
\item $\lVert\mathbf{u}\rVert=6,\theta_u=60^\circ$;
$\lVert\mathbf{v}\rVert=10,\theta_v=120^\circ$;
$\lVert\mathbf{w}\rVert=8,\theta_w=270^\circ$.
\item Three unit vectors at $0^\circ,120^\circ,240^\circ$.
\item $\mathbf{u}=2\hat{i}+3\hat{j}$,
$\mathbf{v}=-5\hat{i}+2\hat{j}$,
$\mathbf{w}=-\hat{j}$.
\item $\mathbf{u}=\hat{i}$, $\mathbf{v}=-\hat{j}$,
$\mathbf{w}=\hat{i}-\hat{j}$.
\end{enumerate}
\item Add
\[
2\hat{i}+3\hat{j}-\hat{k},\quad
\hat{i}+2\hat{k},\quad
\hat{i}+\hat{j}-3\hat{k},
\]
and find the resultant magnitude and direction cosines.
\item Given $A=(-1,2,-1)$, $B=(-3,6,6)$, $C=(4,3,1)$,
$D=(0,0,2)$, find the length and direction cosines of
$\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}$.
\item Using the same points: find the point dividing $AB$ in the ratio $2:1$;
find the centroid of $A,B,C$; and show that the side midpoints of the skew
quadrilateral $ABCD$ form a parallelogram.
\item Prove
\[
\lVert\mathbf{u}+\mathbf{v}\rVert^2
=\lVert\mathbf{u}\rVert^2+\lVert\mathbf{v}\rVert^2
+2\lVert\mathbf{u}\rVert\lVert\mathbf{v}\rVert\cos\theta.
\]
\item Deduce
\[
\cos\theta=\frac{u_xv_x+u_yv_y+u_zv_z}
{\lVert\mathbf{u}\rVert\lVert\mathbf{v}\rVert}.
\]
\item Find the Cartesian equations of the line through
$A=(-2,0,2)$ and $B=(2,1,-3)$.
\end{enumerate}
\section*{Modern notation references}
The notation and terminology in this modernized article follow standard
present-day mechanics and vector-analysis usage, particularly:
\begin{enumerate}
\item J. R. Taylor, \emph{Classical Mechanics}, University Science Books, 2005.
\item D. Kleppner and R. Kolenkow, \emph{An Introduction to Mechanics}, 2nd ed., Cambridge University Press, 2014.
\item H. Goldstein, C. Poole, and J. Safko, \emph{Classical Mechanics}, 3rd ed., Addison--Wesley, 2002.
\end{enumerate}
\section*{Source}
This article is a modernized restatement of the corresponding Public \PMlinkescape{Domain} article in
Louis Brand, \emph{Vectorial Mechanics}, John Wiley \& Sons, New York, 1930,
Chapter I, ``Vector Algebra.'' The original 1930 edition is the source basis.