E7.2 Vectors in two dimensions

Learning objectives

E7.2.1Describe a translation using a vector represented by x y J L KK N P OO,• Describe a translation using a vector represented by x y J L KK N P OO, AB or a.E7.2.2Add and subtract vectors.• Add and subtract vectors.E7.2.3Multiply a vector by a scalar• Multiply a vector by a scalar. Vectors will be printed as AB or a.

A two-dimensional vector records a directed movement: its top component is the horizontal change and its bottom component is the vertical change. The same vector can be written as a column vector, a directed segment such as $\overrightarrow{AB}$ , or a bold letter such as $\mathbf a$ .

\overrightarrow{AB}=B-A=\begin{pmatrix}x_B-x_A\y_B-y_A\end{pmatrix}

Component	Positive	Negative
top	right	left
bottom	up	down

For $A=(6,4)$ and $B=(2,7)$ , $\overrightarrow{AB}=\begin{pmatrix}2-6\\7-4\end{pmatrix}=\begin{pmatrix}-4\\3\end{pmatrix}$ . This means 4 units left and 3 units up. Conversely, if $A=(4,1)$ and $\overrightarrow{AB}=\begin{pmatrix}-3\\1\end{pmatrix}$ , then $B=(1,2)$ .

Always subtract end minus start. Reversing the direction reverses both signs: $\overrightarrow{BA}=-\overrightarrow{AB}$ . This objective describes direction and components; calculating vector magnitude belongs to E7.3.

Vectors add by combining their horizontal changes and their vertical changes separately. Geometrically, addition joins movements head-to-tail, so a route through an intermediate point has the same resultant as the direct route.

\begin{pmatrix}a\b\end{pmatrix}+\begin{pmatrix}c\d\end{pmatrix}=\begin{pmatrix}a+c\b+d\end{pmatrix},\qquad\begin{pmatrix}a\b\end{pmatrix}-\begin{pmatrix}c\d\end{pmatrix}=\begin{pmatrix}a-c\b-d\end{pmatrix}

For points $A,B,C$ , the head-to-tail route law is $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ . Subtracting a vector means adding its reverse: $\mathbf a-\mathbf b=\mathbf a+(-\mathbf b)$ .

Let $\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}-5\\4\end{pmatrix}$ . Then $\mathbf a+\mathbf b=\begin{pmatrix}-2\\2\end{pmatrix}$ , while $\mathbf a-\mathbf b=\begin{pmatrix}8\\-6\end{pmatrix}$ . The different second result comes from subtracting both components of $\mathbf b$ .

Keep corresponding components aligned; do not add the top of one vector to the bottom of another. Route order matters when directed segments are named, because $\overrightarrow{AB}$ and $\overrightarrow{BA}$ have opposite signs.

Multiplying a vector by a scalar $k$ multiplies every component by the same number. The result stays parallel to the original vector: a positive scalar keeps its direction, a negative scalar reverses it, and zero gives the zero vector.

k\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}ka\kb\end{pmatrix}

Scalar $k$	Direction relative to original	Movement scale
$k>0$	same	multiplied by $k$
$k<0$	opposite	multiplied by $\|k\|$
$k=0$	no direction	zero movement

If $\mathbf v=\begin{pmatrix}-3\\7\end{pmatrix}$ , then $2\mathbf v=\begin{pmatrix}-6\\14\end{pmatrix}$ and $-\tfrac12\mathbf v=\begin{pmatrix}\tfrac32\\-\tfrac72\end{pmatrix}$ . Both components change by the same factor, so the vectors remain parallel.

Apply the scalar to both components. Adding $k$ to each component is not scalar multiplication. Notation such as $3\overrightarrow{AB}$ means three times the directed vector from $A$ to $B$ ; it does not mean a new point named $3A$ or $3B$ .

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—E7.3 Magnitude of a vectorRead

E7.2 Vectors in two dimensions

Learning objectives

Describe directed movement with a vector

Add and subtract vectors component by component

Multiply a vector by a scalar

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