5.1 Vectors

Syllabus
2017
Topic
5.1
Level
Higher

Learning objectives

Recognise magnitude and direction in a vector

A vector describes a movement with both magnitude (size) and direction. Two vectors are equal only when both properties match, even if they start at different points.

Quantity What matters
scalar size only
vector size and direction
equal vectors same size and same direction

Read an arrow from its tail to its head: the arrow length represents magnitude and the arrowhead fixes direction. Reversing the arrow produces the negative vector.

Parallel arrows are not automatically equal: they may point oppositely or have different magnitudes.

Read vector and column-vector notation

The vector from OO to AA is written OA\overrightarrow{OA} and may be named a\mathbf a. A column vector (xy)\begin{pmatrix}x\\y\end{pmatrix} means move xx horizontally and yy vertically.

Component Positive Negative
top, xx right left
bottom, yy up down

Order matters: AO=OA=a\overrightarrow{AO}=-\overrightarrow{OA}=-\mathbf a. Translate a diagram into notation by naming the start point first and the end point second.

The entries of a column vector are displacements, not the coordinates of its endpoint unless the vector starts at the origin.

Multiply a vector by a scalar

Multiplying a vector by a scalar kk multiplies every component by kk. Its magnitude is multiplied by k|k|; a negative scalar also reverses its direction.

Scalar kk Effect on vector
k>1k>1 same direction, longer
0<k<10<k<1 same direction, shorter
k=0k=0 zero vector
k<0k<0 reversed direction, scaled by k|k|

If a=(32)\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}, then 2a=(64)-2\mathbf a=\begin{pmatrix}-6\\4\end{pmatrix}.

Do not multiply only one component; the scalar acts on the whole vector.

Add and subtract vectors

Add vectors by joining movements head-to-tail or by adding corresponding components. Subtracting a vector means adding its reverse: ab=a+(b)\mathbf a-\mathbf b=\mathbf a+(-\mathbf b).

Operation Column rule
addition (x1y1)+(x2y2)=(x1+x2y1+y2)\begin{pmatrix}x_1\\y_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\end{pmatrix}=\begin{pmatrix}x_1+x_2\\y_1+y_2\end{pmatrix}
subtraction subtract top from top and bottom from bottom
route AC=AB+BC\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

(53)(24)=(71)\begin{pmatrix}5\\3\end{pmatrix}-\begin{pmatrix}-2\\4\end{pmatrix}=\begin{pmatrix}7\\-1\end{pmatrix}: brackets protect the signs.

A route must connect head-to-tail. If an arrow points the wrong way, reverse it and change its sign before combining.

Calculate the modulus of a vector

For v=(xy)\mathbf v=\begin{pmatrix}x\\y\end{pmatrix}, its modulus is the non-negative length v=x2+y2|\mathbf v|=\sqrt{x^2+y^2}, found using Pythagoras' theorem.

Step Action
1 identify horizontal and vertical components
2 square both components
3 add the squares
4 take the positive square root

(912)=92+122=15\left|\begin{pmatrix}9\\12\end{pmatrix}\right|=\sqrt{9^2+12^2}=15.

The modulus is a scalar, so it has no direction and cannot be negative. Squaring a negative component makes a positive contribution.

Find a resultant vector

A resultant is the single vector with the same overall effect as two or more successive vectors. Follow a continuous route and add every directed segment.

Route fact Vector equation
OABCO\to A\to B\to C OC=OA+AB+BC\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}
reverse a segment BA=AB\overrightarrow{BA}=-\overrightarrow{AB}
closed route vector sum is 0\mathbf0

Choose a route whose start and finish match the required resultant, rewrite every segment in the stated base vectors, then collect the coefficients of each base vector.

Do not add undirected lengths or rely on the visual angle of a diagram; resultant calculations use directed vector equations.

Prove geometrical facts with vectors

A vector proof translates each geometric condition into an exact vector relation. Equal vectors establish equal directed sides; non-zero scalar multiples establish parallel lines, and a specified fraction locates a division point.

Geometric fact Vector evidence
midpoint MM of ABAB AM=12AB\overrightarrow{AM}=\tfrac12\overrightarrow{AB}
parallel lines one direction vector is a non-zero scalar multiple of the other
same point by two routes the two position-vector expressions are equal
collinear points their connecting vectors are scalar multiples

Example: if OC=13a\overrightarrow{OC}=\tfrac13\mathbf a and OD=13b\overrightarrow{OD}=\tfrac13\mathbf b, then CD=13(ba)\overrightarrow{CD}=\tfrac13(\mathbf b-\mathbf a). Since AB=ba\overrightarrow{AB}=\mathbf b-\mathbf a, CDCD is parallel to ABAB.

Finishing with an expression is not a proof. State the geometric conclusion justified by the equality or scalar-multiple relationship, and exclude the zero-vector case when asserting a direction.