Polar form of a vector
Learn the polar form of a vector for Year 11 Specialist Mathematics in Queensland (QCAA). Polar form records a vector as its magnitude and its direction, the natural way to describe velocities, forces and displacements.
You will learn to convert between polar form and component form, find a vector’s magnitude and direction angle, and keep the angle in the correct quadrant — skills that underpin the algebra of vectors and mechanics later in the course.
Theory
Polar form writes a vector as \([r,\theta]\) — its magnitude \(r\) and its direction \(\theta\) — in Year 11 Specialist Mathematics (QCAA, Queensland). This page shows how to convert between polar form and component (Cartesian) form \(x\mathbf{i}+y\mathbf{j}\), and how to keep the direction angle in the correct quadrant.
A vector in the plane has both a magnitude (how long it is) and a direction (which way it points). Polar form records exactly these two facts, writing the vector as \([r,\theta]\).
Here \(r=|\mathbf{v}|\) is the magnitude, always \(r\ge 0\), and \(\theta\) is the direction angle, measured anticlockwise from the positive \(x\)-axis. For example, \([6,90^\circ]\) is a vector of length \(6\) pointing straight up.
The same vector can be written in component (Cartesian) form as \(\mathbf{v}=x\mathbf{i}+y\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the axes. The components \(x\) and \(y\) are the vector's horizontal and vertical shadows on the axes.
In navigation a direction is often given as a bearing, measured clockwise from north. A bearing must be resolved into east and north components rather than read straight off the \(x\)-axis convention.
To go from components to polar form, find the magnitude with Pythagoras and the direction with the tangent ratio:
To go from polar form to components, resolve along each axis:
Converting between the two forms
- Sketch the vector from the origin so you can see which quadrant it lies in.
- Components to polar: compute \(r=\sqrt{x^2+y^2}\), then a reference angle \(\tan^{-1}\left|\dfrac{y}{x}\right|\), and adjust it into the correct quadrant to get \(\theta\).
- Polar to components: compute \(x=r\cos\theta\) and \(y=r\sin\theta\), keeping exact surds where possible.
- State the answer in the form asked — \([r,\theta]\) for polar, or \(x\mathbf{i}+y\mathbf{j}\) for components.
Find the magnitude with Pythagoras:
| \(r\) | \(=\) | \(\sqrt{5^2+12^2}\) |
| \(=\) | \(\sqrt{25+144}\) | |
| \(=\) | \(\sqrt{169}\) | |
| \(=\) | \(13\) |
Both components are positive, so the angle is in the first quadrant:
| \(\tan\theta\) | \(=\) | \(\dfrac{12}{5}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\dfrac{12}{5}\) |
| \(\approx\) | \(67^\circ\) |
Assemble the polar form:
| \(\mathbf{v}\) | \(=\) | \([13,67^\circ]\) |
\(\mathbf{v}=[13,67^\circ]\).
Resolve along the \(x\)-axis with \(x=r\cos\theta\):
| \(x\) | \(=\) | \(8\cos 60^\circ\) |
| \(=\) | \(8\times\dfrac{1}{2}\) | |
| \(=\) | \(4\) |
Resolve along the \(y\)-axis with \(y=r\sin\theta\):
| \(y\) | \(=\) | \(8\sin 60^\circ\) |
| \(=\) | \(8\times\dfrac{\sqrt{3}}{2}\) | |
| \(=\) | \(4\sqrt{3}\) |
\(\mathbf{v}=4\mathbf{i}+4\sqrt{3}\,\mathbf{j}\).
Find the magnitude:
| \(r\) | \(=\) | \(\sqrt{(-1)^2+(\sqrt{3})^2}\) |
| \(=\) | \(\sqrt{1+3}\) | |
| \(=\) | \(\sqrt{4}\) | |
| \(=\) | \(2\) |
Since \(x<0\) and \(y>0\), the vector is in the second quadrant. Use \(\cos\theta\) and \(\sin\theta\) to place it:
| \(\cos\theta\) | \(=\) | \(\dfrac{-1}{2},\ \sin\theta=\dfrac{\sqrt{3}}{2}\) |
| \(\theta\) | \(=\) | \(120^\circ\) |
\(\mathbf{v}=[2,120^\circ]\).
Resolve each vector into components:
| \([4,0^\circ]\) | \(=\) | \((4,\ 0)\) |
| \([4,120^\circ]\) | \(=\) | \((4\cos 120^\circ,\ 4\sin 120^\circ)\) |
| \(=\) | \((-2,\ 2\sqrt{3})\) |
Add the components:
| \(\text{sum}\) | \(=\) | \((4+(-2),\ 0+2\sqrt{3})\) |
| \(=\) | \((2,\ 2\sqrt{3})\) |
Convert the resultant back to polar form:
| \(r\) | \(=\) | \(\sqrt{2^2+(2\sqrt{3})^2}\) |
| \(=\) | \(\sqrt{4+12}\) | |
| \(=\) | \(4\) | |
| \(\tan\theta\) | \(=\) | \(\dfrac{2\sqrt{3}}{2}=\sqrt{3}\) |
| \(\theta\) | \(=\) | \(60^\circ\) |
\([4,0^\circ]+[4,120^\circ]=[4,60^\circ]\).
Common pitfalls
Frequently asked questions
What is the polar form of a vector?
It writes a vector as \([r,\theta]\), where \(r\) is the magnitude (length) and \(\theta\) is the direction angle measured anticlockwise from the positive \(x\)-axis.
How do you convert a vector from component form to polar form?
Find the magnitude \(r=\sqrt{x^2+y^2}\), then the direction from \(\tan\theta=\dfrac{y}{x}\), adjusting for the quadrant, and write \([r,\theta]\).
How do you convert from polar form to components?
Use \(x=r\cos\theta\) and \(y=r\sin\theta\); then \(\mathbf{v}=x\mathbf{i}+y\mathbf{j}\).
Why does my calculator give the wrong direction angle?
Because \(\tan^{-1}\) only returns angles between \(-90^\circ\) and \(90^\circ\). If the vector points left (\(x<0\)), add \(180^\circ\) to the reference angle to land in the correct quadrant.
What is the difference between a direction angle and a bearing?
A direction angle is measured anticlockwise from the positive \(x\)-axis; a bearing is measured clockwise from north. They are different reference directions, so convert carefully.
Is the magnitude of a vector ever negative?
No. The magnitude \(r=|\mathbf{v}|\) is a length, so \(r\ge 0\). A negative scalar multiple changes the direction, not the sign of the magnitude.