A-level Mathematics

10.1 Vectors in two and three dimensions — Learn

Prerequisites: plotting points on axes; subtracting whole numbers and negative numbers; using Pythagoras' theorem.

By the end of this section you can: explain what a vector is; write a vector as a directed line segment, as a column and in i\mathbf{i}, j\mathbf{j}, k\mathbf{k} form; recognise equal, negative and parallel vectors; work with vectors in two dimensions and in three dimensions.

1. Scalars and vectors

Some quantities are fixed by a single number. A mass of 55 kg and a temperature of 2020°C are examples. A number of this kind is called a scalar.

Other quantities need a direction as well as a size. A force of 55 N pushes somewhere. A velocity of 55 m/s points somewhere. A quantity with both a size and a direction is called a vector. The size of a vector is its magnitude.

Displacement is a vector. "Walk 33 km" and "walk 33 km north" are different instructions.

2. Directed line segments and notation

A vector is drawn as an arrow. The length of the arrow is the magnitude. The head of the arrow shows the direction. The arrow can be moved anywhere on the page without changing the vector, because a free vector is fixed only by its length and direction.

Several notations describe the same vector:

The unit vectors i\mathbf{i} and j\mathbf{j} have length 11. The vector i\mathbf{i} points along the positive xx-axis. The vector j\mathbf{j} points along the positive yy-axis.

In the column (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix}, the top number is the xx-component and the bottom number is the yy-component. The column (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix} and the vector 3i+4j3\mathbf{i} + 4\mathbf{j} are the same vector.

Example 1. Write the displacement from A(1,2)A(1, 2) to B(4,6)B(4, 6) as a column and in i\mathbf{i}, j\mathbf{j} form.

Subtract the coordinates of AA from those of BB:

AB⃗=(4−16−2)=(34)=3i+4j.\vec{AB} = \begin{pmatrix} 4 - 1 \\ 6 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 3\mathbf{i} + 4\mathbf{j}.

The vector moves 33 units in the xx-direction and 44 units in the yy-direction.

Example 2. Write the displacement from B(4,6)B(4, 6) to A(1,2)A(1, 2).

BA⃗=(1−42−6)=(−3−4)=−3i−4j.\vec{BA} = \begin{pmatrix} 1 - 4 \\ 2 - 6 \end{pmatrix} = \begin{pmatrix} -3 \\ -4 \end{pmatrix} = -3\mathbf{i} - 4\mathbf{j}.

Changing the order of the letters reverses every sign: BA⃗=−AB⃗\vec{BA} = -\vec{AB}.

3. Equal, negative and parallel vectors

Two vectors are equal when they have the same magnitude and the same direction. They do not need to start at the same point. In components, every component must match.

The vector −a-\mathbf{a} has the same magnitude as a\mathbf{a} but the opposite direction.

Two vectors are parallel when one is a scalar multiple of the other. If b=ka\mathbf{b} = k\mathbf{a} for some number kk, then b\mathbf{b} is parallel to a\mathbf{a}. A positive kk keeps the direction. A negative kk reverses it.

Example 3. Show that a=3i−2j\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} and PQ⃗\vec{PQ}, where P(1,5)P(1, 5) and Q(4,3)Q(4, 3), are equal.

PQ⃗=(4−13−5)=(3−2)=3i−2j=a.\vec{PQ} = \begin{pmatrix} 4 - 1 \\ 3 - 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} = 3\mathbf{i} - 2\mathbf{j} = \mathbf{a}.

Both components match, so the two vectors are equal.

Example 4. Show that u=2i−j\mathbf{u} = 2\mathbf{i} - \mathbf{j} and v=6i−3j\mathbf{v} = 6\mathbf{i} - 3\mathbf{j} are parallel.

Compare the components. Each component of v\mathbf{v} is 33 times the matching component of u\mathbf{u}:

v=6i−3j=3(2i−j)=3u.\mathbf{v} = 6\mathbf{i} - 3\mathbf{j} = 3(2\mathbf{i} - \mathbf{j}) = 3\mathbf{u}.

The multiplier is 33, which is positive, so v\mathbf{v} is parallel to u\mathbf{u} and points the same way.

4. Vectors in three dimensions

A point in three dimensions has three coordinates, written (x,y,z)(x, y, z). A vector in three dimensions has three components:

a=a1i+a2j+a3k=(a1a2a3).\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}.

The unit vector k\mathbf{k} points along the positive zz-axis. Its direction is often called upwards.

Example 5. Write the displacement from A(1,2,−1)A(1, 2, -1) to B(4,−1,3)B(4, -1, 3).

AB⃗=(4−1−1−23−(−1))=(3−34)=3i−3j+4k.\vec{AB} = \begin{pmatrix} 4 - 1 \\ -1 - 2 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ 4 \end{pmatrix} = 3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}.

Example 6. A vector is a=2i+j−5k\mathbf{a} = 2\mathbf{i} + \mathbf{j} - 5\mathbf{k}. Write down −2a-2\mathbf{a}.

Multiply each component by −2-2:

−2a=−4i−2j+10k.-2\mathbf{a} = -4\mathbf{i} - 2\mathbf{j} + 10\mathbf{k}.

The new vector is parallel to a\mathbf{a} and points the opposite way.

5. Try it: one vector, drawn anywhere

The sliders set the two components of a\mathbf{a} and the point where a\mathbf{a} starts. The second arrow is the multiple kak\mathbf{a}, drawn from the origin.

Predict first. Before you touch a control, predict what happens to the column vector of a\mathbf{a} when you move its starting point across the grid. Then predict which value of kk makes the second arrow equal to a\mathbf{a}. Write down both answers.

Test. Move the two start sliders and watch the column vector and the length. Then set the components to 33 and 44, and set kk to 11, then to −1-1, then to 22. Watch the second arrow each time.

Explain. Write two sentences. The first says why two arrows that start at different points are still the same vector. The second says why a negative kk gives an arrow that is parallel to a\mathbf{a} and points the other way.

Watch out. Three slips are common on this page.

A point is not a vector. The pair (3,4)(3, 4) is a place when it names a point, and a displacement when it names a vector. Read the sentence to decide which meaning is wanted.

The order of the components. A column vector keeps xx on top, yy below it, and zz at the bottom. Writing the components in the wrong order changes the vector, even though the numbers look the same.

The direction of the subtraction. To find AB⃗\vec{AB}, subtract the coordinates of AA from the coordinates of BB. Subtracting the other way gives BA⃗\vec{BA}, which points the opposite way. Re-check: substitute the first point back. If you add your answer to AA, you must reach BB.

You should now be able to

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