Addition of Two Vectors

If $\bar a$ and $\bar b$ are two vectors, their sum (resultant) $\bar a+\bar b$ can be constructed by either

of two equivalent laws.

Parallelogram Law. Draw $\overrightarrow{AB}=\bar a$ and $\overrightarrow{AD}=\bar b$ as two adjacent

sides of a parallelogram $ABCD$ starting from the common point $A$. Then $\bar a+\bar b$ is the diagonal

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$ through $A$.

Triangle Law. Draw $\overrightarrow{AB}=\bar a$ and then, starting exactly where the first arrow ended,

$\overrightarrow{BC}=\bar b$; then $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\bar a+\bar b$ is the third side of the triangle, closing it back to the start. Applying the triangle law twice inside the

same parallelogram ($\triangle ABC$ and $\triangle ADC$) re-derives the parallelogram law, confirming the two

are the same rule seen two ways.

Subtraction. $\bar a-\bar b$ is defined as $\bar a+(-\bar b)$: reverse the arrow for $\bar b$ and add by

the triangle law. Concretely, if $\overrightarrow{AB}=\bar a,\ \overrightarrow{BC}=\bar b$, and $D$ is the

point with $\overrightarrow{BD}=-\overrightarrow{BC}=-\bar b$, then

$\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\bar a-\bar b$.

Rules to remember. Use the parallelogram law when two vectors act simultaneously from the same point

(e.g. two velocities/forces acting together); use the triangle law when they act one after another (e.g.

consecutive displacements). Adding a vector to its own negative always gives $\bar 0$; in any triangle $ABC$,

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\bar 0$ because the path returns to its start.

Addition of vectors is commutative ($\bar a+\bar b=\bar b+\bar a$) and associative

($\bar a+(\bar b+\bar c)=(\bar a+\bar b)+\bar c$), $\bar 0$ is the additive identity ($\bar a+\bar 0=\bar a$),

$-\bar a$ is the additive inverse ($\bar a+(-\bar a)=\bar 0$), and for scalars $m,n$: $(m+n)\bar a=m\bar a+n\bar a$, $m(\bar a+\bar b)=m\bar a+m\bar b$, $m(n\bar a)=(mn)\bar a=n(m\bar a)$. Finally, the Triangle Inequality $|\bar a+\bar b|\le|\bar a|+|\bar b|$ follows because one side of a triangle is never longer than

the sum of the other two; and any two vectors $\bar a,\bar b$ span a plane in which $\bar a+\bar b$ and

$\bar a-\bar b$ both lie.

Polygon (extended) law. For vectors $\bar a=\overrightarrow{PQ},\ \bar b=\overrightarrow{QR},\ \bar c=\overrightarrow{RS},\ \bar d=\overrightarrow{ST}$ placed tip-to-tail around a polygon, repeated use of

the triangle law collapses the whole sum to the single vector spanning start to end:

$\bar a+\bar b+\bar c+\bar d=\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}+\overrightarrow{ST} =\overrightarrow{PT}$.

Worked examples.

• In a polygon $KLNM$ with $\overrightarrow{KL}=\bar a,\overrightarrow{LN}=\bar b,\overrightarrow{NM}=\bar c,\overrightarrow{KT}=\bar d$, the polygon law gives, e.g., $\overrightarrow{KM}=\overrightarrow{KL} +\overrightarrow{LN}+\overrightarrow{NM}=\bar a+\bar b+\bar c$, and similarly $\overrightarrow{LT}=\overrightarrow{KT}-\overrightarrow{KL}=\bar d-\bar a$ (found from the triangle law in $\triangle KLT$), while $\overrightarrow{MT}=\bar d-\bar a-\bar b-\bar c$ is built by chaining several triangle-law steps together.
• Midpoint theorem by vectors. If $M,N$ are midpoints of sides $AB,AC$ of $\triangle ABC$, then $\overrightarrow{AM}=\tfrac12\overrightarrow{AB}$ and $\overrightarrow{AN}=\tfrac12\overrightarrow{AC}$, so …