5.1 - Vectors, resultants and magnitude

Describe movements using vectors, combine directed journeys and calculate the length of a direct displacement. You will connect arrows, bold letters and column notation, then use signed components to keep track of where a journey ends.

Direction, magnitude and notation (Higher)

Can you travel a nonzero distance yet have zero displacement? Imagine walking 5 units east, then 5 units west. You have certainly moved, but your finishing position needs careful description. Vectors let us separate the journey from its overall effect.

You already know

Coordinates locate points. Subtracting a starting coordinate from a finishing coordinate gives a signed change. Pythagoras' theorem finds the hypotenuse of a right-angled triangle.

Moving 5 units east and moving 5 units north have the same size but different directions. A number giving only the size cannot distinguish them.

Vector

A vector has both magnitude and direction. Its magnitude, also called its modulus, is its length. A scalar has size without a direction, such as a length alone.

The directed segment $\overrightarrow{AB}$ goes from $A$ to $B$ . A vector may also be written $\mathbf a$ in bold, or $\vec a$ with an arrow. Equal vectors have the same magnitude and direction; their arrows may start at different places. Equal lengths alone do not make vectors equal.

In a column vector, the top entry is horizontal movement and the bottom entry is vertical movement. Right and up are positive. Thus $\begin{pmatrix}-3\\2\end{pmatrix}$ means 3 left and 2 up. It takes $(0,0)$ to $(-3,2)$ , but takes $(4,1)$ to $(1,3)$ . The vector is a movement, whereas each coordinate pair names a point.

Worked example 1: from endpoints to a vector

Question. Find $\overrightarrow{AB}$ for $A(-2,1)$ and $B(4,9)$ .

Plan. Subtract start from finish in each coordinate.

\begin{aligned} \Delta x&=4-(-2)=6 &&\text{horizontal change}\\ \Delta y&=9-1=8 &&\text{vertical change}\\ \overrightarrow{AB}&=\begin{pmatrix}6\\8\end{pmatrix} &&\text{write the changes as a column} \end{aligned}

Why this order? The arrow starts at $A$ , so its changes must take $A$ to $B$ .

Answer. $\overrightarrow{AB}=\begin{pmatrix}6\\8\end{pmatrix}$ .

Check. $(-2+6,1+8)=(4,9)=B$ .

Change only the direction of the journey. The endpoints stay the same, but the arrow now runs backwards.

Worked example 2: reversing the arrow

Question. For the same points, find $\overrightarrow{BA}$ .

Plan. Use $B$ as the start and $A$ as the finish.

\begin{aligned} \Delta x&=-2-4=-6 &&\text{finish minus start}\\ \Delta y&=1-9=-8 &&\text{finish minus start}\\ \overrightarrow{BA}&=\begin{pmatrix}-6\\-8\end{pmatrix}=-\overrightarrow{AB} &&\text{reverse both movements} \end{aligned}

Why both signs? Reversing the journey reverses both its horizontal and vertical movements.

Answer. $\overrightarrow{BA}=\begin{pmatrix}-6\\-8\end{pmatrix}$ .

Check. $(4-6,9-8)=(-2,1)=A$ .

Always read the order of the letters before calculating. This becomes especially useful when a longer route contains backwards arrows.

Watch out

The line $\overrightarrow{AB}=A-B$ reverses the intended direction. Use finishing coordinates minus starting coordinates: $\overrightarrow{AB}=B-A$ .

Try a reversed journey with a new pair of points.

Scalar multiples (Higher)

Repeating the same movement twice doubles both components. Multiplying by a negative scalar also reverses the direction. The length is multiplied by the scalar's absolute value, so a vector never has a negative magnitude.

Scalar multiplication

For any scalar $k$ ,

k\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}kx\\ky\end{pmatrix},\qquad |k\mathbf v|=|k|\,|\mathbf v|

Not in the Higher formulae sheet: learn it.

For a nonzero vector, a positive factor keeps its direction and a negative factor reverses it. A positive factor between 0 and 1 shortens it. The zero vector $\mathbf0=\begin{pmatrix}0\\0\end{pmatrix}$ has magnitude zero and no unique direction.

Worked example 3: a positive multiple

Question. For $\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$ , find $2\mathbf a$ and describe its direction and length compared with $\mathbf a$ .

Plan. Multiply both components by 2 and interpret the positive factor.

\begin{aligned} 2\mathbf a&=\begin{pmatrix}2\times3\\2\times(-2)\end{pmatrix} &&\text{scale each component}\\ &=\begin{pmatrix}6\\-4\end{pmatrix} &&\text{calculate} \end{aligned}

Why the same factor twice? Scaling an arrow changes both movements in the same proportion.

Answer. $\begin{pmatrix}6\\-4\end{pmatrix}$ , in the same direction and twice as long.

Check. Halving both components recovers $\mathbf a$ .

Now change only the sign of the scalar. This keeps the doubled length but reverses the arrow.

Worked example 4: a negative multiple

Question. For the same $\mathbf a$ , find $-2\mathbf a$ and compare its direction and length with $\mathbf a$ .

Plan. Multiply both components by $-2$ .

\begin{aligned} -2\mathbf a&=\begin{pmatrix}-2\times3\\-2\times(-2)\end{pmatrix} &&\text{scale both entries}\\ &=\begin{pmatrix}-6\\4\end{pmatrix} &&\text{negative times negative is positive} \end{aligned}

Why does the direction reverse? Right and down become left and up.

Answer. $\begin{pmatrix}-6\\4\end{pmatrix}$ , opposite in direction and twice as long.

Check. Its squared length is $36+16=52=4(9+4)$ , so its length is twice $\sqrt{13}$ .

To recognise scalar multiples, the same factor must fit both components. For example, $\begin{pmatrix}6\\-3\end{pmatrix}$ is not twice $\begin{pmatrix}3\\-2\end{pmatrix}$ . Check a zero component directly rather than dividing by it.

Watch out

$-2\begin{pmatrix}3\\-2\end{pmatrix}=\begin{pmatrix}-6\\-4\end{pmatrix}$ is wrong: the lower entry also needs multiplication by $-2$ . It should be $4$ .

Change the scalar once more, this time using a fraction.

Adding and subtracting vectors (Higher)

Addition joins movements head to tail: place the next arrow's tail at the previous arrow's head. Add the horizontal entries together and the vertical entries together. The sum goes from the original start to the final finish. Addition can be done in either order because the total changes are the same.

Vector addition and subtraction

\begin{pmatrix}p\\q\end{pmatrix}+\begin{pmatrix}r\\s\end{pmatrix}=\begin{pmatrix}p+r\\q+s\end{pmatrix},\qquad \mathbf a-\mathbf b=\mathbf a+(-\mathbf b)

Not in the Higher formulae sheet: learn it.

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Number

Algebra

Shape, space and measure

Handling data

5.1 - Vectors, resultants and magnitude

Direction, magnitude and notation (Higher)

Scalar multiples (Higher)

Adding and subtracting vectors (Higher)

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