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Year 11 Specialist (Unit 1 & 2) Vectors in the Plane

Introduction to vectors

20 practice questions 0 video lessons Theory + worked examples

Get started with vectors in Year 11 Specialist Mathematics for Queensland (QCAA). A vector carries both size and direction — like displacement, velocity or force — while a scalar such as distance or speed has size only.

You will learn to draw a vector as a directed line segment, find its magnitude and direction, use vector notation and equality, and combine vectors with the triangle rule — the foundation for everything that follows in vectors.

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Theory

A vector has both size and direction, unlike a scalar which has size only. This page introduces vectors in Year 11 Specialist Mathematics (QCAA, Queensland): how to draw a vector as a directed line segment, its magnitude and direction, vector notation, equality, scalar multiples, and the triangle rule for combining vectors.

A scalar is a quantity with size only, such as distance, speed or mass. A vector has both a magnitude (size) and a direction. Displacement, velocity and force are vectors: distance is how far you travelled, but displacement also says which way.

A vector is drawn as a directed line segment — an arrow whose length shows the magnitude and whose arrowhead shows the direction. The vector from \(A\) to \(B\) is written \(\overrightarrow{AB}\), or with a bold letter \(\mathbf{a}\), and its magnitude is \(|\mathbf{a}|\). A unit vector has magnitude \(1\).

Two vectors are equal when they have the same magnitude and the same direction — where they start does not matter. A scalar multiple \(k\mathbf{a}\) points the same way as \(\mathbf{a}\) when \(k>0\) and the opposite way when \(k<0\); its length is \(|k|\) times the length of \(\mathbf{a}\). The negative \(-\mathbf{a}\) is the same length as \(\mathbf{a}\) but reversed, and the zero vector \(\mathbf{0}\) has no direction.

Vectors are added with the triangle rule: place them head to tail, and the resultant runs from the tail of the first to the head of the last. Reversing an arrow negates the vector, so a difference \(\mathbf{b}-\mathbf{a}\) can be read as \(-\mathbf{a}\) followed by \(\mathbf{b}\). Any vector can be written as a combination of sums, differences and scalar multiples of others.

A vector as a directed line segment An arrow from point A to point B labelled bold a; a dashed horizontal shows the direction angle theta and the arrow length is its magnitude. θ a |a| A B
A vector \(\mathbf{a}=\overrightarrow{AB}\): the arrow's length is its magnitude \(|\mathbf{a}|\) and \(\theta\) gives its direction.
Triangle rule for adding two vectors Vector a from A to B and vector b from B to C are placed head to tail; the resultant a plus b runs straight from A to C. a b a+b A B C
Triangle rule: place \(\mathbf{a}\) then \(\mathbf{b}\) head to tail; the resultant \(\mathbf{a}+\mathbf{b}\) closes the triangle.

For a vector with perpendicular parts of \(v_1\) horizontally and \(v_2\) vertically, the magnitude comes from Pythagoras:

\[ |\mathbf{v}| = \sqrt{v_1^{\,2}+v_2^{\,2}} \]
|v|=v12+v22

Multiplying a vector by a scalar \(k\) scales its magnitude by \(|k|\):

\[ |k\mathbf{a}| = |k|\,|\mathbf{a}| \]
|ka|=|k||a|

The triangle rule adds vectors head to tail, and reversing an arrow negates it:

\[ \overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}, \qquad \overrightarrow{BA} = -\overrightarrow{AB} \]
AB+BC=AC
Magnitude ignores the sign. Because \(|k\mathbf{a}|=|k|\,|\mathbf{a}|\), the vectors \(3\mathbf{a}\) and \(-3\mathbf{a}\) have the same length; only their direction differs.

Expressing a vector as a combination

  1. Read the diagram: note which vectors are given (e.g. \(\overrightarrow{AB}=\mathbf{a}\)) and which one you must find.
  2. Trace a path from the start point to the end point along arrows you know, going head to tail.
  3. Reverse where needed: if you must travel against an arrow, use its negative, so \(\overrightarrow{BA}=-\mathbf{a}\).
  4. Add and simplify: collect the scalar multiples, e.g. \(-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}\); take magnitudes with Pythagoras only at the end.
Example 1 — Magnitude of a displacement
A drone flies \(5\) m due east, then \(12\) m due north. Find the magnitude of its displacement from the start.
Solution

Displacement is the straight-line vector; the two legs are perpendicular, so use Pythagoras:

\(|\mathbf{d}|\)\(=\)\(\sqrt{5^2+12^2}\)
\(=\)\(\sqrt{25+144}\)
\(=\)\(\sqrt{169}\)
\(=\)\(13\)

The displacement has magnitude \(13\) m.

|d|=169=13
Example 2 — Magnitude of a scalar multiple
The vector \(\mathbf{a}\) has magnitude \(|\mathbf{a}|=6\). Find \(|-3\mathbf{a}|\), and state how \(-3\mathbf{a}\) points relative to \(\mathbf{a}\).
Solution

Take the size of the scalar times the magnitude; the sign only reverses direction:

\(|-3\mathbf{a}|\)\(=\)\(|-3|\,|\mathbf{a}|\)
\(=\)\(3\times 6\)
\(=\)\(18\)

Read off the direction from the sign of the scalar:

\(-3\)\(<\)\(0\)
\(\Rightarrow\)\(=\)\(\text{opposite direction to }\mathbf{a}\)

\(|-3\mathbf{a}|=18\), pointing the opposite way to \(\mathbf{a}\).

|3a|=18
Example 3 — Express a side of a triangle
In triangle \(OAB\), \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). Express \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
Solution

Trace a path from \(A\) to \(B\) through \(O\), reversing the arrow \(OA\):

\(\overrightarrow{AB}\)\(=\)\(\overrightarrow{AO}+\overrightarrow{OB}\)
\(=\)\(-\overrightarrow{OA}+\overrightarrow{OB}\)
\(=\)\(-\mathbf{a}+\mathbf{b}\)
\(=\)\(\mathbf{b}-\mathbf{a}\)

\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\).

Triangle OAB with OA=a, OB=b Arrow a from O to A and arrow b from O to B; the side AB is the difference b minus a. a b b−a O A B
Example 4 — Scalar multiple along a line
Points \(P\), \(Q\), \(R\) lie on a straight line with \(\overrightarrow{PQ}=\mathbf{a}\) and \(\overrightarrow{QR}=3\mathbf{a}\). If \(|\mathbf{a}|=2\) m, find \(\overrightarrow{PR}\) and its magnitude.
Solution

Add the two displacements head to tail, then collect the scalar multiples:

\(\overrightarrow{PR}\)\(=\)\(\overrightarrow{PQ}+\overrightarrow{QR}\)
\(=\)\(\mathbf{a}+3\mathbf{a}\)
\(=\)\(4\mathbf{a}\)

Now take the magnitude, using \(|k\mathbf{a}|=|k|\,|\mathbf{a}|\):

\(|\overrightarrow{PR}|\)\(=\)\(|4\mathbf{a}|\)
\(=\)\(4\times 2\)
\(=\)\(8\)

\(\overrightarrow{PR}=4\mathbf{a}\) with magnitude \(8\) m.

|PR|=8

Common pitfalls

Confusing distance with displacement. Distance is a scalar (add the legs); displacement is a vector (the straight line, found with Pythagoras). Walking \(6\) m then \(8\) m is \(14\) m of distance but a displacement of magnitude \(10\) m.
Letting a minus sign change the magnitude. A magnitude is never negative: \(|-3\mathbf{a}|=3|\mathbf{a}|\), not \(-3|\mathbf{a}|\). The sign only tells you the direction.
Adding \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) directly. The triangle rule needs vectors head to tail. To get \(\overrightarrow{BC}\) go \(B\to A\to C\), i.e. \(-\mathbf{a}+\mathbf{b}\), not \(\mathbf{a}+\mathbf{b}\).
Thinking equal vectors must start at the same point. Vectors are equal when they share magnitude and direction; two arrows in different places can be the same vector.

Frequently asked questions

What is the difference between a scalar and a vector?

A scalar has size only (distance, speed, mass). A vector has both size and direction (displacement, velocity, force).

How do you find the magnitude of a vector?

If its perpendicular parts are \(v_1\) and \(v_2\), the magnitude is \(|\mathbf{v}|=\sqrt{v_1^{\,2}+v_2^{\,2}}\) by Pythagoras.

What does multiplying a vector by a scalar do?

It scales the length by \(|k|\) and keeps the direction if \(k>0\) or reverses it if \(k<0\); so \(|k\mathbf{a}|=|k|\,|\mathbf{a}|\).

When are two vectors equal?

When they have the same magnitude and the same direction. Their starting points do not need to match.

What is the triangle rule for adding vectors?

Place the vectors head to tail; the resultant runs from the tail of the first to the head of the last, so \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).

How do you write a vector in the opposite direction?

Negate it. \(-\mathbf{a}\) has the same length as \(\mathbf{a}\) but points the opposite way, and \(\overrightarrow{BA}=-\overrightarrow{AB}\).