Vector projections
Master vector projections in Year 11 VCE Specialist Mathematics. A projection measures how much of one vector points along another: the scalar projection gives that amount as a number, and the vector projection gives it as a vector. 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 use the scalar product to find both projections, the perpendicular component, and resolve a vector into parallel and perpendicular parts — skills behind later work and force problems.
Theory
A projection answers "how much of one vector points along another". In Year 11 Specialist Mathematics the scalar projection gives that amount as a number and the vector projection gives it as a vector along the second direction. This page defines both, shows how to find the perpendicular component, and works through resolving a vector into components.
Given two vectors \(\mathbf{a}\) and \(\mathbf{b}\), a projection measures how much of \(\mathbf{a}\) lies in the direction of \(\mathbf{b}\). It is built entirely from the scalar product \(\mathbf{a}\cdot\mathbf{b}\) and the magnitude \(|\mathbf{b}|\).
The scalar projection (also called the component) of \(\mathbf{a}\) in the direction of \(\mathbf{b}\) is the number \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\). It is the signed length of the shadow \(\mathbf{a}\) casts onto the line of \(\mathbf{b}\).
The vector projection of \(\mathbf{a}\) onto \(\mathbf{b}\), written \(\text{proj}_{\mathbf{b}}\,\mathbf{a}\), is that shadow as a vector: \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^{2}}\,\mathbf{b}\). It always points along \(\mathbf{b}\) (or the opposite way).
The leftover part, \(\mathbf{a}-\text{proj}_{\mathbf{b}}\,\mathbf{a}\), is the component of \(\mathbf{a}\) perpendicular to \(\mathbf{b}\). Adding the parallel and perpendicular components back together gives \(\mathbf{a}\); this is what "resolving a vector into components" means. A negative scalar projection tells you the angle between the vectors is obtuse.
For the scalar projection (component) of \(\mathbf{a}\) in the direction of \(\mathbf{b}\):
For the vector projection of \(\mathbf{a}\) onto \(\mathbf{b}\):
For the component perpendicular to \(\mathbf{b}\):
How to project one vector onto another
- Find the scalar product \(\mathbf{a}\cdot\mathbf{b}\) by multiplying matching components and adding.
- Find \(|\mathbf{b}|\) (or \(|\mathbf{b}|^{2}\)): use \(|\mathbf{b}|=\sqrt{b_1^{2}+b_2^{2}}\). For a vector projection you only need \(|\mathbf{b}|^{2}=b_1^{2}+b_2^{2}\), so no surd appears.
- Divide: for the scalar projection compute \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\); for the vector projection multiply \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^{2}}\) by \(\mathbf{b}\).
- Perpendicular part if asked: subtract the vector projection from \(\mathbf{a}\), then check it by confirming its scalar product with \(\mathbf{b}\) is \(0\).
Scalar product first, then divide by \(|\mathbf{b}|\):
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((6)(3)+(8)(4)\) |
| \(=\) | \(18+32\) | |
| \(=\) | \(50\) | |
| \(|\mathbf{b}|\) | \(=\) | \(\sqrt{3^{2}+4^{2}}\) |
| \(=\) | \(\sqrt{25}=5\) | |
| \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\) | \(=\) | \(\dfrac{50}{5}\) |
| \(=\) | \(10\) |
The scalar projection is \(10\).
The denominator is a surd, so keep the answer as an exact fraction:
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((5)(1)+(2)(3)\) |
| \(=\) | \(5+6\) | |
| \(=\) | \(11\) | |
| \(|\mathbf{b}|\) | \(=\) | \(\sqrt{1^{2}+3^{2}}\) |
| \(=\) | \(\sqrt{10}\) | |
| \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\) | \(=\) | \(\dfrac{11}{\sqrt{10}}\) |
The scalar projection is \(\dfrac{11}{\sqrt{10}}\).
Use \(|\mathbf{b}|^{2}\) (no surd needed) then multiply the fraction by \(\mathbf{b}\):
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((1)(1)+(3)(1)\) |
| \(=\) | \(4\) | |
| \(|\mathbf{b}|^{2}\) | \(=\) | \(1^{2}+1^{2}\) |
| \(=\) | \(2\) | |
| \(\text{proj}_{\mathbf{b}}\,\mathbf{a}\) | \(=\) | \(\dfrac{4}{2}(\mathbf{i}+\mathbf{j})\) |
| \(=\) | \(2(\mathbf{i}+\mathbf{j})\) | |
| \(=\) | \(2\mathbf{i}+2\mathbf{j}\) |
The vector projection is \(2\mathbf{i}+2\mathbf{j}\).
Parallel part is the vector projection:
| \(\mathbf{a}\cdot\mathbf{b}\) | \(=\) | \((2)(4)+(6)(2)\) |
| \(=\) | \(8+12\) | |
| \(=\) | \(20\) | |
| \(|\mathbf{b}|^{2}\) | \(=\) | \(4^{2}+2^{2}\) |
| \(=\) | \(20\) | |
| \(\text{proj}_{\mathbf{b}}\,\mathbf{a}\) | \(=\) | \(\dfrac{20}{20}(4\mathbf{i}+2\mathbf{j})\) |
| \(=\) | \(4\mathbf{i}+2\mathbf{j}\) |
Perpendicular part is \(\mathbf{a}\) minus that projection:
| \(\mathbf{a}-\text{proj}_{\mathbf{b}}\,\mathbf{a}\) | \(=\) | \((2\mathbf{i}+6\mathbf{j})-(4\mathbf{i}+2\mathbf{j})\) |
| \(=\) | \(-2\mathbf{i}+4\mathbf{j}\) |
Check the perpendicular part really is perpendicular to \(\mathbf{b}\):
| \((-2\mathbf{i}+4\mathbf{j})\cdot\mathbf{b}\) | \(=\) | \((-2)(4)+(4)(2)\) |
| \(=\) | \(-8+8\) | |
| \(=\) | \(0\ \checkmark\) |
Parallel component \(4\mathbf{i}+2\mathbf{j}\); perpendicular component \(-2\mathbf{i}+4\mathbf{j}\).
Common pitfalls
Frequently asked questions
What is the difference between a scalar projection and a vector projection?
The scalar projection is a single number, \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\), giving the signed length along \(\mathbf{b}\). The vector projection is a vector, \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^{2}}\,\mathbf{b}\), pointing along \(\mathbf{b}\).
Why does the vector projection divide by the magnitude squared?
One factor of \(|\mathbf{b}|\) turns \(\mathbf{a}\cdot\mathbf{b}\) into the scalar projection; the second turns \(\mathbf{b}\) into the unit vector \(\dfrac{\mathbf{b}}{|\mathbf{b}|}\). Together they give \(|\mathbf{b}|^{2}\).
What does a negative scalar projection mean?
It means the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is obtuse, so \(\mathbf{a}\) points partly against \(\mathbf{b}\). The vector projection then points opposite to \(\mathbf{b}\).
How do I find the component of a vector perpendicular to another?
Subtract the vector projection from the original vector: \(\mathbf{a}-\text{proj}_{\mathbf{b}}\,\mathbf{a}\). You can check it by confirming its scalar product with \(\mathbf{b}\) is zero.
What happens if the two vectors are perpendicular?
Then \(\mathbf{a}\cdot\mathbf{b}=0\), so both the scalar projection and the vector projection are zero — \(\mathbf{a}\) casts no shadow along \(\mathbf{b}\).
What does resolving a vector into components mean?
It means writing the vector as the sum of two perpendicular parts: one parallel to a chosen direction (the vector projection) and one perpendicular to it. The two parts add back to the original vector.