Polar form of a vector
Master the polar form of a vector in Year 11 VCE Specialist Mathematics. Polar form records a vector as its magnitude and its direction, the natural way to describe velocities, forces and displacements. 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 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. 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.