Components of vectors
Master the components of vectors in Year 11 VCE Specialist Mathematics. A plane vector is written along the perpendicular unit vectors \(\mathbf{i}\) and \(\mathbf{j}\), which turns every calculation into simple arithmetic on the components. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Vectors in the plane topic of Unit 2.
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.
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. 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.
For a vector \(\mathbf{a}=v_1\mathbf{i}+v_2\mathbf{j}\), the magnitude (length) is found by Pythagoras:
The direction (angle from the positive \(x\)-axis) comes from the tangent ratio:
The unit vector in the direction of \(\mathbf{a}\) is \(\mathbf{a}\) divided by its magnitude:
Working with components
- Read off the components: write the vector as \(v_1\mathbf{i}+v_2\mathbf{j}\), keeping the sign of each component.
- Magnitude: square each component, add, and take the square root, \(|\mathbf{a}|=\sqrt{v_1^{\,2}+v_2^{\,2}}\).
- Direction: use \(\tan\theta=\dfrac{v_2}{v_1}\) and take the inverse tangent for the angle from the positive \(x\)-axis.
- Unit vector: divide each component by the magnitude, \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\).
Square each component, add, then take the square root:
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{7^2+24^2}\) |
| \(=\) | \(\sqrt{49+576}\) | |
| \(=\) | \(\sqrt{625}\) | |
| \(=\) | \(25\) |
\(|\mathbf{a}|=25\).
First find the magnitude:
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{(-4)^2+3^2}\) |
| \(=\) | \(\sqrt{16+9}\) | |
| \(=\) | \(\sqrt{25}\) | |
| \(=\) | \(5\) |
Then divide each component by the magnitude:
| \(\hat{\mathbf{a}}\) | \(=\) | \(\dfrac{1}{5}(-4\mathbf{i}+3\mathbf{j})\) |
| \(=\) | \(-\dfrac{4}{5}\mathbf{i}+\dfrac{3}{5}\mathbf{j}\) |
\(\hat{\mathbf{a}}=-\dfrac{4}{5}\mathbf{i}+\dfrac{3}{5}\mathbf{j}\).
Use the tangent ratio, then take the inverse tangent:
| \(\tan\theta\) | \(=\) | \(\dfrac{v_2}{v_1}\) |
| \(=\) | \(\dfrac{7}{3}\) | |
| \(\theta\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{7}{3}\right)\) |
| \(=\) | \(66.8^\circ\) | |
| \(\approx\) | \(67^\circ\) |
\(\theta\approx 67^\circ\).
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
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 \(B\).