Vector Projections
A ramp makes an angle with the floor. A force acts at an angle. In both cases, you need to "shadow" one vector onto another — that shadow is a projection. The scalar projection $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$ gives the signed length; the vector projection $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$ gives direction too. This lesson unpacks both.
A 10 N force acts at 60° above the horizontal. Without using a formula — estimate the horizontal component (the "shadow" of the force on the ground). Write your reasoning.
Every projection question has two forms of the answer: the scalar projection (a signed length) and the vector projection (a vector in the direction of $\mathbf{b}$).
To project $\mathbf{a}$ onto $\mathbf{b}$:
Scalar: $\text{comp}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$
Vector: $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$
The vector projection is always parallel to $\mathbf{b}$. The scalar projection is its signed length.
Key facts
- Scalar projection: $\text{comp}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$
- Vector projection: $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$
- The vector projection is always parallel to $\mathbf{b}$
Concepts
- Why the scalar projection can be negative (when $\theta > 90°$)
- The relationship $\text{proj}_{\mathbf{b}}\mathbf{a} = (\text{comp}_{\mathbf{b}}\mathbf{a})\,\hat{\mathbf{b}}$
- Physical interpretation: projections decompose forces along and perpendicular to surfaces
Skills
- Calculate scalar and vector projections in 2D and 3D
- Find the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$
- Apply projections to real-world problems involving forces and displacement
The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is the signed length of the shadow of $\mathbf{a}$ on the line of $\mathbf{b}$:
It is positive when $\theta$ is acute (the shadow is in the direction of $\mathbf{b}$) and negative when $\theta$ is obtuse (the shadow is opposite to $\mathbf{b}$).
Force example: A 10 N force at 60° to the horizontal. Taking $\mathbf{b} = \begin{pmatrix}1\\0\end{pmatrix}$ (horizontal unit vector), the scalar projection is $10\cos 60° = 10 \times \frac{1}{2} = 5$ N. So the horizontal component is 5 N — check your hook estimate!
The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is the signed length of the shadow of $\mathbf{a}$ on the line of $\mathbf{b}$:
Pause — copy the scalar projection formula: $\text{comp}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ with the geometric interpretation (signed shadow length) into your book.
Quick check: The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is given by:
We just saw the scalar projection of $\vec{a}$ onto $\vec{b}$: $\text{comp}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ — a signed length. That raises a question: the vector projection is a vector in the direction of $\hat{b}$; how do you build it from the scalar projection, and what is the formula in terms of $\vec{b}$? This card answers it → multiply the scalar projection by $\hat{b}=\vec{b}/|\vec{b}|$: $\text{proj}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$.
The vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is a vector in the direction of $\mathbf{b}$ with magnitude equal to the scalar projection:
The denominator $|\mathbf{b}|^2$ comes from multiplying the scalar projection $\dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$ by the unit vector $\hat{\mathbf{b}} = \dfrac{\mathbf{b}}{|\mathbf{b}|}$.
Orthogonal decomposition: Any vector $\mathbf{a}$ can be split into a part parallel to $\mathbf{b}$ and a part perpendicular to $\mathbf{b}$:
Example: Find the vector projection of $\mathbf{a} = \begin{pmatrix}3\\4\end{pmatrix}$ onto $\mathbf{b} = \begin{pmatrix}1\\2\end{pmatrix}$.
$\mathbf{a} \cdot \mathbf{b} = 3(1) + 4(2) = 11$
$|\mathbf{b}|^2 = 1^2 + 2^2 = 5$
$\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{11}{5}\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}11/5\\22/5\end{pmatrix}$
The vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is a vector in the direction of $\mathbf{b}$ with magnitude equal to the scalar projection:
Pause — copy the vector projection formula: $\text{proj}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}=\frac{\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\vec{b}$ into your book.
Did you get this? True or false: the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is always parallel to $\mathbf{b}$.
Worked examples · 3 in a row, reveal as you go
Find the scalar and vector projections of $\mathbf{a} = \begin{pmatrix}4\\3\end{pmatrix}$ onto $\mathbf{b} = \begin{pmatrix}3\\0\end{pmatrix}$.
Express $\mathbf{a} = \begin{pmatrix}5\\2\end{pmatrix}$ as the sum of a vector parallel to $\mathbf{b} = \begin{pmatrix}2\\1\end{pmatrix}$ and a vector perpendicular to $\mathbf{b}$.
Find the vector projection of $\mathbf{u} = \begin{pmatrix}1\\2\\3\end{pmatrix}$ onto $\mathbf{v} = \begin{pmatrix}1\\0\\1\end{pmatrix}$.
Fill the gap: For $\mathbf{a} = \begin{pmatrix}6\\0\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}3\\4\end{pmatrix}$, the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ uses $|\mathbf{b}|^2$ (not $|\mathbf{b}|$) in the denominator.
Activities · practice with the ideas
Find the scalar and vector projections of $\mathbf{a} = \begin{pmatrix}3\\4\end{pmatrix}$ onto $\mathbf{b} = \begin{pmatrix}1\\0\end{pmatrix}$.
Find $\text{proj}_{\mathbf{v}}\mathbf{u}$ where $\mathbf{u} = \begin{pmatrix}2\\6\end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix}1\\3\end{pmatrix}$.
Decompose $\mathbf{a} = \begin{pmatrix}4\\1\end{pmatrix}$ into components parallel and perpendicular to $\mathbf{b} = \begin{pmatrix}2\\3\end{pmatrix}$. Verify the perpendicular component is orthogonal to $\mathbf{b}$.
Find the scalar projection of $\mathbf{p} = \begin{pmatrix}1\\-3\\2\end{pmatrix}$ onto $\mathbf{q} = \begin{pmatrix}2\\1\\2\end{pmatrix}$.
A 20 N force acts at 30° to a slope. Find the component of the force along the slope.
Odd one out: Three of these statements about projections are correct. Which one is FALSE?
Earlier you estimated the horizontal component of a 10 N force at 60° to the horizontal.
The answer is $10\cos 60° = 10 \times \tfrac{1}{2} = \mathbf{5}$ N. This is the scalar projection of the force vector onto the horizontal unit vector $\hat{\mathbf{i}}$. The vertical component is $10\sin 60° = 5\sqrt{3} \approx 8.66$ N — larger than the horizontal, because the force is closer to vertical than horizontal at 60°.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the scalar and vector projections of $\mathbf{a} = \begin{pmatrix}6\\2\end{pmatrix}$ onto $\mathbf{b} = \begin{pmatrix}1\\2\end{pmatrix}$. (2 marks)
Q2. Express $\mathbf{a} = \begin{pmatrix}3\\-1\end{pmatrix}$ as the sum of a vector parallel to $\mathbf{b} = \begin{pmatrix}1\\1\end{pmatrix}$ and a vector perpendicular to $\mathbf{b}$. Verify your answer. (3 marks)
Q3. A displacement vector $\mathbf{d} = \begin{pmatrix}4\\2\\-4\end{pmatrix}$ metres represents a journey. Find the component of $\mathbf{d}$ in the direction of $\mathbf{u} = \begin{pmatrix}1\\2\\2\end{pmatrix}$, then find the distance from the path of $\mathbf{u}$ to the tip of $\mathbf{d}$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. Scalar = 3, vector = $\begin{pmatrix}3\\0\end{pmatrix}$ · 2. $\mathbf{u} \cdot \mathbf{v} = 20$, $|\mathbf{v}|^2 = 10$, proj $= 2\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}2\\6\end{pmatrix}$ (note $\mathbf{u}$ itself!) · 3. proj $= \dfrac{11}{13}\begin{pmatrix}2\\3\end{pmatrix}$, perp $= \begin{pmatrix}4 - 22/13\\1-33/13\end{pmatrix}$; check dot product $= 0$ · 4. $\mathbf{p} \cdot \mathbf{q} = 2-3+4 = 3$, $|\mathbf{q}| = 3$, scalar $= 1$ · 5. $20\cos30° = 10\sqrt{3} \approx 17.3$ N
Q1 (2 marks): $\mathbf{a} \cdot \mathbf{b} = 6+4 = 10$ [M1]. $|\mathbf{b}|^2 = 5$. Scalar $= \dfrac{10}{\sqrt{5}} = 2\sqrt{5}$ [A1]. Vector $= \dfrac{10}{5}\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}2\\4\end{pmatrix}$ [A1 if asked for both].
Q2 (3 marks): $\mathbf{a} \cdot \mathbf{b} = 2$, $|\mathbf{b}|^2 = 2$. Proj $= \begin{pmatrix}1\\1\end{pmatrix}$ [M1]. Perp $= \begin{pmatrix}3\\-1\end{pmatrix} - \begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}2\\-2\end{pmatrix}$ [A1]. Check: $\begin{pmatrix}2\\-2\end{pmatrix} \cdot \begin{pmatrix}1\\1\end{pmatrix} = 2-2 = 0$ ✓ [A1].
Q3 (3 marks): $\mathbf{d} \cdot \mathbf{u} = 4+4-8 = 0$; scalar projection $= 0$ [M1], so $\mathbf{d}$ is perpendicular to $\mathbf{u}$. Vector projection $= \mathbf{0}$ [A1]. Perpendicular distance $= |\mathbf{d}| = \sqrt{16+4+16} = 6$ m [A1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering vector projection questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.