Vector projections
Learn vector projections for Year 11 Specialist Mathematics in Queensland (QCAA). 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.
You will 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 (QCAA, Queensland) 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.