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

Components of vectors

20 practice questions 0 video lessons Theory + worked examples

Master the components of vectors for Year 11 Specialist Mathematics in Queensland (QCAA). A plane vector is written along the perpendicular unit vectors i and j, which turns every calculation into simple arithmetic on the components.

You will learn to find a vector's magnitude and direction, work out its unit vector, and find the vector between two points — the component-form foundation for vector algebra, the dot product and applications later in the course.

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Theory

Components of vectors express a plane vector along the perpendicular unit vectors i and j as \(\mathbf{a}=v_1\mathbf{i}+v_2\mathbf{j}\), in Year 11 Specialist Mathematics (QCAA, Queensland). From the components you find a vector's magnitude, its direction, and its unit vector. This page shows each calculation with full worked examples.

Every vector in the plane can be built from two unit vectors: \(\mathbf{i}\) points one unit along the positive \(x\)-axis and \(\mathbf{j}\) points one unit along the positive \(y\)-axis. They are perpendicular, so any vector is a combination of them.

Writing \(\mathbf{a}=v_1\mathbf{i}+v_2\mathbf{j}\) is component (Cartesian) form. The numbers \(v_1\) and \(v_2\) are the components of the vector; \(v_1\) is the horizontal part and \(v_2\) the vertical part. This is the same information as the column vector \(\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\).

The magnitude \(|\mathbf{a}|\) is the length of the vector, and the direction is the angle \(\theta\) it makes with the positive \(x\)-axis. A unit vector \(\hat{\mathbf{a}}\) is a vector of length \(1\) that points the same way as \(\mathbf{a}\).

The vector between two points \(A\) and \(B\) is \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), found by subtracting the position vectors component by component.

Magnitude of a vector from its components The vector a = 6i + 8j drawn from the origin; its horizontal component 6 and vertical component 8 form a right triangle whose hypotenuse is the magnitude 10. x y a 6 8 |a| = 10
Magnitude: the components of \(\mathbf{a}=6\mathbf{i}+8\mathbf{j}\) form a right triangle, so \(|\mathbf{a}|=\sqrt{6^2+8^2}=10\).
Unit vector in the direction of a vector The vector a = 9i + 12j of length 15, and the unit vector a-hat of length 1 pointing in the same direction from the origin. x y a |a| = 15 â (length 1)
Unit vector: \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\) keeps the direction of \(\mathbf{a}=9\mathbf{i}+12\mathbf{j}\) but has length \(1\).

For a vector \(\mathbf{a}=v_1\mathbf{i}+v_2\mathbf{j}\), the magnitude (length) is found by Pythagoras:

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

The direction (angle from the positive \(x\)-axis) comes from the tangent ratio:

\[ \tan\theta = \dfrac{v_2}{v_1} \]
tanθ=v2v1

The unit vector in the direction of \(\mathbf{a}\) is \(\mathbf{a}\) divided by its magnitude:

\[ \hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} = \dfrac{v_1}{|\mathbf{a}|}\mathbf{i}+\dfrac{v_2}{|\mathbf{a}|}\mathbf{j} \]
a^=a|a|
Square, add, then root. The magnitude is never \(v_1+v_2\); you must square each component, add, then take the square root. A unit vector always has magnitude \(1\), which is a quick check on your answer.

Working with components

  1. Read off the components: write the vector as \(v_1\mathbf{i}+v_2\mathbf{j}\), keeping the sign of each component.
  2. Magnitude: square each component, add, and take the square root, \(|\mathbf{a}|=\sqrt{v_1^{\,2}+v_2^{\,2}}\).
  3. Direction: use \(\tan\theta=\dfrac{v_2}{v_1}\) and take the inverse tangent for the angle from the positive \(x\)-axis.
  4. Unit vector: divide each component by the magnitude, \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\).
Example 1 — Magnitude
Find the magnitude of \(\mathbf{a}=7\mathbf{i}+24\mathbf{j}\).
Solution

Square each component, add, then take the square root:

\(|\mathbf{a}|\)\(=\)\(\sqrt{7^2+24^2}\)
\(=\)\(\sqrt{49+576}\)
\(=\)\(\sqrt{625}\)
\(=\)\(25\)

\(|\mathbf{a}|=25\).

Example 2 — Unit vector
Find the unit vector in the direction of \(\mathbf{a}=-4\mathbf{i}+3\mathbf{j}\).
Solution

First find the magnitude:

\(|\mathbf{a}|\)\(=\)\(\sqrt{(-4)^2+3^2}\)
\(=\)\(\sqrt{16+9}\)
\(=\)\(\sqrt{25}\)
\(=\)\(5\)

Then divide each component by the magnitude:

\(\hat{\mathbf{a}}\)\(=\)\(\dfrac{1}{5}(-4\mathbf{i}+3\mathbf{j})\)
\(=\)\(-\dfrac{4}{5}\mathbf{i}+\dfrac{3}{5}\mathbf{j}\)

\(\hat{\mathbf{a}}=-\dfrac{4}{5}\mathbf{i}+\dfrac{3}{5}\mathbf{j}\).

Example 3 — Direction
Find the angle \(\theta\) that \(\mathbf{v}=3\mathbf{i}+7\mathbf{j}\) makes with the positive \(x\)-axis, to the nearest degree.
Solution

Use the tangent ratio, then take the inverse tangent:

\(\tan\theta\)\(=\)\(\dfrac{v_2}{v_1}\)
\(=\)\(\dfrac{7}{3}\)
\(\theta\)\(=\)\(\tan^{-1}\!\left(\dfrac{7}{3}\right)\)
\(=\)\(66.8^\circ\)
\(\approx\)\(67^\circ\)

\(\theta\approx 67^\circ\).

Example 4 — Vector between two points
Points \(A(-1,2)\) and \(B(4,14)\) are given. Find \(\overrightarrow{AB}\) and its length.
Solution

Subtract the position vectors, matching components:

\(\overrightarrow{AB}\)\(=\)\(\mathbf{b}-\mathbf{a}\)
\(=\)\((4-(-1))\mathbf{i}+(14-2)\mathbf{j}\)
\(=\)\(5\mathbf{i}+12\mathbf{j}\)

Then take the magnitude of the result:

\(|\overrightarrow{AB}|\)\(=\)\(\sqrt{5^2+12^2}\)
\(=\)\(\sqrt{169}\)
\(=\)\(13\)

\(\overrightarrow{AB}=5\mathbf{i}+12\mathbf{j}\), with length \(13\).

Common pitfalls

Adding the components for magnitude. The magnitude is not \(v_1+v_2\). Square each component, add, then take the square root: \(|\mathbf{a}|=\sqrt{v_1^{\,2}+v_2^{\,2}}\).
Dropping a negative sign. When you square a negative component, keep it inside brackets, \((-4)^2=16\), not \(-4^2\). The sign of each component also fixes which quadrant the direction lies in.
Only halving the unit vector. To make a unit vector you divide both components by the magnitude, not just one. Check that the result has length \(1\).

Frequently asked questions

How do you find the magnitude of a vector in component form?

Square each component, add the squares, then take the square root: for \(\mathbf{a}=v_1\mathbf{i}+v_2\mathbf{j}\), \(|\mathbf{a}|=\sqrt{v_1^{\,2}+v_2^{\,2}}\).

What are the unit vectors i and j?

\(\mathbf{i}\) is a vector of length \(1\) along the positive \(x\)-axis and \(\mathbf{j}\) is a vector of length \(1\) along the positive \(y\)-axis. They are perpendicular, so any plane vector is \(v_1\mathbf{i}+v_2\mathbf{j}\).

How do you find a unit vector?

Divide the vector by its magnitude: \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\). Both components are divided by \(|\mathbf{a}|\), and the result always has length \(1\).

How do you find the direction of a vector?

Use \(\tan\theta=\dfrac{v_2}{v_1}\) and take the inverse tangent, where \(\theta\) is the angle from the positive \(x\)-axis. Check the signs of the components to place the angle in the correct quadrant.

How do you find the vector from point A to point B?

Subtract the position vectors: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), taking the difference of the \(x\)-components and of the \(y\)-components. Its length is the distance from \(A\) to \(B\).

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