E8.2 - Adding and scaling vectors (Extended)

E8.2 - Adding and scaling vectors (Extended)

Two different routes can finish at the same point. E8.2 uses directed segments, column vectors and bold lower-case letters to describe their displacements. You will add and subtract vectors, then multiply them by numbers to change an arrow's length or reverse its direction.

Vector notation

If one arrow starts at (1,2)(1,2) and ends at (4,4)(4,4), while another starts at (−2,0)(-2,0) and ends at (1,2)(1,2), do they represent the same vector? Both move 3 right and 2 up. A vector records that movement, not one compulsory starting position.

Vector notation

A vector has magnitude and direction. The directed segment AB→\overrightarrow{AB} means the displacement from A to B. A bold lower-case letter such as a\mathbf{a} can name the same vector:

a=AB→=(32).\mathbf{a}=\overrightarrow{AB}=\begin{pmatrix}3\\2\end{pmatrix}.

Equal vectors have the same magnitude and direction, even when their arrows are drawn in different positions.

Use the arrow over two letters to distinguish a displacement from the ordinary length AB. Read its start and finish in that order. A coordinate pair locates a point; a column vector describes a movement between points.

Worked example 1: a directed segment as a column

Question. A(−2,1)A(-2,1) and B(3,4)B(3,4) are points. Write AB→\overrightarrow{AB} as a column vector and name it a\mathbf{a}.

Plan. Find the signed changes from A to B, horizontally first.

horizontal change=3−(−2)=5vertical change=4−1=3\begin{aligned} \text{horizontal change}&=3-(-2)=5\\ \text{vertical change}&=4-1=3 \end{aligned}

Answer. a=AB→=(53)\mathbf{a}=\overrightarrow{AB}=\begin{pmatrix}5\\3\end{pmatrix}.

Check. Start at A and move 5 right and 3 up: (−2+5,1+3)=(3,4)=B(-2+5,1+3)=(3,4)=B.

Now keep the endpoints but reverse their order. The direction reverses, so both signed movements change.

Worked example 2: reversing a directed segment

Question. For the same points, write BA→\overrightarrow{BA} as a column vector.

Plan. Start at B and find the movement to A.

horizontal change=−2−3=−5vertical change=1−4=−3\begin{aligned} \text{horizontal change}&=-2-3=-5\\ \text{vertical change}&=1-4=-3 \end{aligned}

Answer. BA→=(−5−3)=−a\overrightarrow{BA}=\begin{pmatrix}-5\\-3\end{pmatrix}=-\mathbf{a}.

Check. The journey A to B followed by B to A returns to its start. Its horizontal changes 5+(−5)5+(-5) and vertical changes 3+(−3)3+(-3) are both zero.

Two arrows can have equal length but different directions, so equal length alone does not make equal vectors. Reversing an arrow preserves its length but produces the negative vector. The zero vector, (00)\begin{pmatrix}0\\0\end{pmatrix}, gives no displacement and has no specified direction.

Watch out

AB→\overrightarrow{AB} and BA→\overrightarrow{BA} are opposites, not alternative names for the same non-zero vector. The endpoint minus the starting point must match the order of the letters.

Use the directed letters rather than the order in which the coordinates happen to be listed.

Addition and subtraction

Place the start of a second vector at the endpoint of the first, without turning or resizing either arrow. The direct arrow from the first start to the final finish represents their sum. This is the head-to-tail rule.

Adding and subtracting components

Combine horizontal components with horizontal components, and vertical with vertical:

(ab)+(cd)=(a+cb+d)(ab)−(cd)=(a−cb−d)\begin{aligned} \begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix} &=\begin{pmatrix}a+c\\b+d\end{pmatrix}\\ \begin{pmatrix}a\\b\end{pmatrix}-\begin{pmatrix}c\\d\end{pmatrix} &=\begin{pmatrix}a-c\\b-d\end{pmatrix} \end{aligned}

Subtracting a vector means adding its opposite. These component rules are not on the supplied formula list for Papers 1–4: learn them.

The component rule follows from the journey. If the first move goes 3 right and the next 1 left, the net horizontal displacement is 2 right. Work out the vertical displacement separately. The sum records the overall displacement, not the distance travelled along the two arrows.

[DIAGRAM: asset_name: A sum and a difference as journeys; asset_slug: vector_sum_difference; caption: For subtraction, reverse the second arrow before placing it head to tail.; description: Two accurate coordinate grids start at O zero comma zero and use a equals three comma two and b equals minus one comma four. Left shows a to three comma two, followed by b to two comma six, with the direct sum arrow from O to two comma six. Right shows a followed by negative b, one right and four down, to four comma minus two, with the direct difference arrow from O. Labelled arrows and pale colours distinguish first, second and resultant vectors.]
For subtraction, reverse the second arrow before placing it head to tail.

Worked example 3: adding two displacements

Question. a=(32)\mathbf{a}=\begin{pmatrix}3\\2\end{pmatrix} and b=(−14)\mathbf{b}=\begin{pmatrix}-1\\4\end{pmatrix}. Find a+b\mathbf{a}+\mathbf{b}.

Plan. Add corresponding signed components, interpreting the result as one journey.

a+b=(3+(−1)2+4)=(26).\mathbf{a}+\mathbf{b} =\begin{pmatrix}3+(-1)\\2+4\end{pmatrix} =\begin{pmatrix}2\\6\end{pmatrix}.

Answer. (26)\begin{pmatrix}2\\6\end{pmatrix}: 2 right and 6 up overall.

Check. Starting at the origin, a\mathbf{a} reaches (3,2)(3,2). Moving 1 left and 4 up then reaches (2,6)(2,6), the endpoint of the direct sum arrow.

Change only the operation next. Subtraction reverses the second vector; it does not reverse the first.

Continue this lesson.

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