\subsection{Scalars, Vectors, and Components}

This subsection introduces the language used throughout Unit 1 for quantities that have magnitude only and for quantities that also have direction.

\dfn{Scalars, vectors, and component form}{A \textbf{scalar} quantity is specified completely by a magnitude. Examples include mass, time, temperature, energy, and speed.

A \textbf{vector} quantity is specified by both a magnitude and a direction. Examples include displacement, velocity, acceleration, and force.

In a chosen Cartesian coordinate system, let $x$, $y$, and $z$ denote coordinates along the $x$-, $y$-, and $z$-axes, and let $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$ denote unit vectors along those axes. If $\vec{v}$ is a vector with scalar components $v_x$, $v_y$, and $v_z$, then its unit-vector decomposition is
\[
\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}.
\]
In two dimensions,
\[
\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath}.
\]
The numbers $v_x$, $v_y$, and $v_z$ are \textbf{components} of $\vec{v}$; they are scalars and can be positive, negative, or zero. The magnitude of $\vec{v}$ is written $|\vec{v}|$.}

\nt{Speed is a scalar, but velocity is a vector. A component such as $v_x$ is a scalar, not a vector by itself. Also, the magnitude $|\vec{v}|$ is not the same thing as a component: $|\vec{v}|\geq 0$, while a component can be negative if the vector points partly in a negative coordinate direction. Likewise, distance is a scalar, while displacement $\Delta \vec{r}$ is a vector. Later in this unit, $\Delta \vec{r}$, $\vec{v}$, and $\vec{a}$ will all be handled with the same component ideas.}

\mprop{Operational rules in components}{Let
\[
\vec{u} = u_x\hat{\imath} + u_y\hat{\jmath} + u_z\hat{k}
\qquad\text{and}\qquad
\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}
\]
be vectors written in the same Cartesian coordinate system.

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item Vector addition and subtraction are done component-by-component:
\[
\vec{u} + \vec{v} = (u_x+v_x)\hat{\imath} + (u_y+v_y)\hat{\jmath} + (u_z+v_z)\hat{k},
\]
\[
\vec{u} - \vec{v} = (u_x-v_x)\hat{\imath} + (u_y-v_y)\hat{\jmath} + (u_z-v_z)\hat{k}.
\]

\item In Cartesian coordinates, the magnitude of a vector comes from its components:
\[
|\vec{v}| = \sqrt{v_x^2+v_y^2}
\qquad\text{in 2D,}
\]
\[
|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2}
\qquad\text{in 3D.}
\]

\item In two dimensions, if $\theta$ is the direction angle of $\vec{v}$ measured from the positive $x$-axis, then
\[
\tan\theta = \frac{v_y}{v_x},
\]
provided $v_x\neq 0$. The signs of $v_x$ and $v_y$ determine the correct quadrant for $\theta$. If $v_x=0$, then the vector points straight along the positive or negative $y$-axis.
\end{enumerate}}

\qs{Worked example}{In an $xy$ coordinate system, the positive $x$-axis points east and the positive $y$-axis points north. A student first moves with displacement
\[
\Delta \vec{r}_1=(3.0\hat{\imath}+4.0\hat{\jmath})\,\text{m}
\]
and then moves with displacement
\[
\Delta \vec{r}_2=(-1.0\hat{\imath}+2.0\hat{\jmath})\,\text{m}.
\]
Let $\Delta \vec{r}_{\text{tot}}$ denote the total displacement, and let $d$ denote the total distance traveled.

\begin{enumerate}[label=(\alph*)]
\item Identify whether $\Delta \vec{r}_{\text{tot}}$, the component $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$, and $d$ are scalars or vectors.
\item Find $\Delta \vec{r}_{\text{tot}}$ in component form.
\item Find $|\Delta \vec{r}_{\text{tot}}|$.
\item Let $\theta$ be the direction of $\Delta \vec{r}_{\text{tot}}$ measured counterclockwise from the positive $x$-axis. Find $\theta$.
\end{enumerate}}
\sol For part (a), $\Delta \vec{r}_{\text{tot}}$ is a vector because it has both magnitude and direction. The quantity $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$ is a scalar because it is one component of a vector. The quantity $d$ is also a scalar because distance gives only a path length.

For part (b), add the two displacements by components:
\[
\Delta \vec{r}_{\text{tot}}=\Delta \vec{r}_1+\Delta \vec{r}_2.
\]
Therefore,
\[
\Delta \vec{r}_{\text{tot}}=(3.0-1.0)\hat{\imath}+(4.0+2.0)\hat{\jmath}.
\]
So,
\[
\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m}.
\]

For part (c), use the magnitude formula in two dimensions:
\[
|\Delta \vec{r}_{\text{tot}}|=\sqrt{(2.0\,\text{m})^2+(6.0\,\text{m})^2}.
\]
Thus,
\[
|\Delta \vec{r}_{\text{tot}}|=\sqrt{40}\,\text{m}=6.32\,\text{m}.
\]

For part (d), use the component ratio. Since both components of $\Delta \vec{r}_{\text{tot}}$ are positive, the vector lies in the first quadrant. Then
\[
\tan\theta=\frac{6.0}{2.0}=3.0.
\]
So,
\[
\theta=\tan^{-1}(3.0)=71.6^\circ.
\]
Therefore, the total displacement is
\[
\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m},
\]
with magnitude $6.32\,\text{m}$ and direction $71.6^\circ$ counterclockwise from the positive $x$-axis.

As a useful comparison, the total distance traveled is
\[
d=|\Delta \vec{r}_1|+|\Delta \vec{r}_2|=5.0\,\text{m}+\sqrt{5}\,\text{m}=7.24\,\text{m},
\]
which is a scalar and is not equal to the magnitude of the total displacement.