Mixed Vector Problems
You have all the tools — dot product, projections, vector lines, magnitude, unit vectors. Now it's time to pick the right one for each problem. This lesson drills multi-technique problems that combine several vector skills in a single question, building the fluency you need under HSC exam conditions.
Given $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j}$, without computing — write down which technique(s) you would use to find: (i) the angle between $\mathbf{a}$ and $\mathbf{b}$, (ii) the projection of $\mathbf{a}$ onto $\mathbf{b}$, and (iii) a unit vector parallel to $\mathbf{a}$.
Mixed problems require you to quickly identify what is being asked before reaching for a formula. Here is the decision map:
Goal: angle between vectors — use the dot product formula $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$.
Goal: projection of $\mathbf{a}$ onto $\mathbf{b}$ — use $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$ (vector) or $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ (scalar).
Goal: unit vector — divide by magnitude: $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$.
Goal: point on a line — use the parametric form $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ and substitute the parameter.
Goal: perpendicularity — show $\mathbf{a}\cdot\mathbf{b} = 0$.
Goal: parallelism — show $\mathbf{a} = k\mathbf{b}$ for some scalar $k$.
Key facts
- Angle: $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$, result in $[0°, 180°]$
- Vector projection: $\text{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$
- Unit vector: $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$, always magnitude 1
Concepts
- How to chain multiple vector operations across a multi-part HSC question
- When a dot product is needed versus a magnitude or direction calculation
- How to use results from earlier parts of a question in later parts
Skills
- Solve 3–4 part mixed vector problems without being told which technique to use
- Identify perpendicular and parallel vectors by inspection or by calculation
- Find the angle, projection, and unit vector for given 2D and 3D vectors
Most HSC vector questions follow a predictable scaffolded structure: early parts establish ingredients (magnitudes, a dot product value) and later parts use them. Recognising this structure saves time.
Typical flow in a 5–6 mark vector question:
- Part (a): Evaluate $\mathbf{a}\cdot\mathbf{b}$ or $|\mathbf{a}|$ — pure arithmetic, 1 mark.
- Part (b): Find the angle between $\mathbf{a}$ and $\mathbf{b}$ — uses the result of (a), 1–2 marks.
- Part (c): Find the projection of one vector onto another — again uses (a), 1–2 marks.
- Part (d): Deduce a geometric property or find an unknown component — 2 marks, requires synthesis.
Also watch for questions that embed a vector problem inside a geometry context — "triangle $OAB$ where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$". The vector translation is: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, the midpoint $M$ has position vector $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$, and the angle at $O$ uses $\cos\angle AOB = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$.
Most HSC vector questions follow a predictable scaffolded structure: early parts establish ingredients (magnitudes, a dot product value) and later parts use them. Recognising this structure saves time.
Pause — copy the HSC scaffolded structure: read what's established in parts (a)-(b) before attempting later parts; identify which formula each part needs into your book.
Quick check: Which expression gives the angle $\theta$ between $\mathbf{a} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{b} = -\mathbf{i} + 3\mathbf{j}$?
We just saw the typical HSC scaffolded structure: part (a) establishes vectors, part (b) uses the dot product for angle or projection, part (c) applies a further result (e.g.\ area, component form). That raises a question: the angle formula and projection formula both use $\vec{a}\cdot\vec{b}$ — when do you compute an angle and when do you compute a projection? This card answers it → angle: divide the dot product by BOTH magnitudes; projection: divide by only ONE magnitude (the one you're projecting onto), or its square for the vector form.
The dot product underpins both the angle formula and the projection formula, but the outputs are different:
For the angle formula, divide the dot product by both magnitudes. For the projection formula, divide by $|\mathbf{b}|^2$ (not $|\mathbf{b}|^2$ times $|\mathbf{a}|$) and multiply back by the direction vector $\mathbf{b}$.
Full worked mini-example: Let $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j}$.
- $\mathbf{a}\cdot\mathbf{b} = 3(1) + 4(-2) = 3 - 8 = -5$
- $|\mathbf{a}| = \sqrt{9+16} = 5$, $|\mathbf{b}| = \sqrt{1+4} = \sqrt{5}$
- Angle: $\cos\theta = \dfrac{-5}{5\sqrt{5}} = \dfrac{-1}{\sqrt{5}}$, so $\theta = \cos^{-1}\!\!\left(\dfrac{-1}{\sqrt{5}}\right) \approx 116.6°$
- Projection of $\mathbf{a}$ onto $\mathbf{b}$: $\dfrac{-5}{(\sqrt{5})^2}(\mathbf{i}-2\mathbf{j}) = \dfrac{-5}{5}(\mathbf{i}-2\mathbf{j}) = -\mathbf{i}+2\mathbf{j}$
The dot product underpins both the angle formula and the projection formula, but the outputs are different:
Pause — copy the angle-vs-projection decision: angle needs $|\vec{a}|\cdot|\vec{b}|$ in denominator; scalar projection needs $|\vec{b}|$; vector projection needs $|\vec{b}|^2$ (or $\vec{b}\cdot\vec{b}$) into your book.
Did you get this? True or false: if $\mathbf{a}\cdot\mathbf{b} = 0$ and both vectors are non-zero, the angle between them is exactly 90°.
Worked examples · 3 in a row, reveal as you go
Let $\mathbf{u} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and $\mathbf{v} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$.
(a) Find the angle between $\mathbf{u}$ and $\mathbf{v}$, correct to the nearest degree.
(b) Find a unit vector in the direction of $\mathbf{u} + \mathbf{v}$.
$|\mathbf{u}| = \sqrt{4+1+4} = 3$, $|\mathbf{v}| = \sqrt{1+4+4} = 3$
$|\mathbf{u}+\mathbf{v}| = \sqrt{9+1+0} = \sqrt{10}$
Unit vector $= \dfrac{1}{\sqrt{10}}(3\mathbf{i}+\mathbf{j}) = \dfrac{3}{\sqrt{10}}\mathbf{i} + \dfrac{1}{\sqrt{10}}\mathbf{j}$
Let $\mathbf{a} = \mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = 4\mathbf{i} + 2\mathbf{j}$.
(a) Find the vector projection of $\mathbf{a}$ onto $\mathbf{b}$.
(b) Hence, find the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$.
$|\mathbf{b}|^2 = 16+4 = 20$
Check: $(-\mathbf{i}+2\mathbf{j})\cdot(4\mathbf{i}+2\mathbf{j}) = -4+4 = 0$ ✓
Vector $\mathbf{p} = 3\mathbf{i} + m\mathbf{j}$ is perpendicular to $\mathbf{q} = 2\mathbf{i} - 3\mathbf{j}$. Find $m$, and hence find the unit vector in the direction of $\mathbf{p}$.
$3(2) + m(-3) = 0 \Rightarrow 6 - 3m = 0 \Rightarrow m = 2$
Fill the gap: If $\mathbf{a} = k\mathbf{i} - 4\mathbf{j}$ is perpendicular to $\mathbf{b} = 2\mathbf{i} + \mathbf{j}$, then $k = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is always parallel to $\mathbf{b}$.
Activities · practice with the ideas
Given $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = 3\mathbf{i} - \mathbf{j}$, find (i) $\mathbf{a}\cdot\mathbf{b}$, (ii) the angle between $\mathbf{a}$ and $\mathbf{b}$ to the nearest degree.
Find the vector projection of $\mathbf{p} = 5\mathbf{i} + 2\mathbf{j}$ onto $\mathbf{q} = 2\mathbf{i} + \mathbf{j}$, and hence find the component of $\mathbf{p}$ perpendicular to $\mathbf{q}$.
Find the value of $t$ such that $\mathbf{a} = t\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ is perpendicular to $\mathbf{b} = 2\mathbf{i} - \mathbf{j} + 4\mathbf{k}$.
Triangle $OAB$ has $\overrightarrow{OA} = 2\mathbf{i} + \mathbf{j}$ and $\overrightarrow{OB} = \mathbf{i} + 3\mathbf{j}$. Find the angle $\angle AOB$ to the nearest degree.
Given $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$, find a unit vector in the direction of $\mathbf{u}$ and a unit vector perpendicular to $\mathbf{u}$ (in 2D there are two choices — state both).
Odd one out: Three of these statements are correct. Which one is NOT?
Earlier you identified which techniques to use for the given vectors $\mathbf{a} = 3\mathbf{i}+4\mathbf{j}$ and $\mathbf{b} = \mathbf{i}-2\mathbf{j}$.
For the angle: $\cos\theta = \dfrac{-5}{5\sqrt{5}} = \dfrac{-1}{\sqrt{5}} \Rightarrow \theta \approx 116.6°$. For the projection of $\mathbf{a}$ onto $\mathbf{b}$: $-\mathbf{i}+2\mathbf{j}$. For a unit vector parallel to $\mathbf{a}$: $\hat{\mathbf{a}} = \tfrac{3}{5}\mathbf{i}+\tfrac{4}{5}\mathbf{j}$. Did your instincts align with the approach?
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. Let $\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$. Find the angle between $\mathbf{a}$ and $\mathbf{b}$, correct to the nearest degree. (2 marks)
Q2. Let $\mathbf{p} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{q} = 4\mathbf{i} - 3\mathbf{j}$.
(a) Show that $\mathbf{p}$ and $\mathbf{q}$ are perpendicular. (1 mark)
(b) Find the unit vector in the direction of $\mathbf{p} - \mathbf{q}$. (2 marks)
Q3. Triangle $OAB$ has position vectors $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ where $|\mathbf{a}| = 3$, $|\mathbf{b}| = 4$ and $\mathbf{a}\cdot\mathbf{b} = 6$. Find the angle $\angle AOB$ to the nearest degree, and hence find the exact length of $AB$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\mathbf{a}\cdot\mathbf{b} = 3-2 = 1$; $|\mathbf{a}|=\sqrt{5}$, $|\mathbf{b}|=\sqrt{10}$; $\cos\theta = \dfrac{1}{\sqrt{50}} = \dfrac{1}{5\sqrt{2}}$; $\theta \approx 82°$.
2. $\mathbf{p}\cdot\mathbf{q} = 10+2=12$; $|\mathbf{q}|^2=5$; $\text{proj} = \tfrac{12}{5}(2\mathbf{i}+\mathbf{j}) = \tfrac{24}{5}\mathbf{i}+\tfrac{12}{5}\mathbf{j}$; perp $= (5-\tfrac{24}{5})\mathbf{i}+(2-\tfrac{12}{5})\mathbf{j} = \tfrac{1}{5}\mathbf{i}-\tfrac{2}{5}\mathbf{j}$. Check: $(\tfrac{1}{5}\mathbf{i}-\tfrac{2}{5}\mathbf{j})\cdot(2\mathbf{i}+\mathbf{j}) = \tfrac{2}{5}-\tfrac{2}{5}=0$ ✓
3. $\mathbf{a}\cdot\mathbf{b} = 2t-3-4=0 \Rightarrow t = \tfrac{7}{2}$
4. $\mathbf{a}\cdot\mathbf{b} = 2+3=5$; $|\mathbf{a}|=\sqrt{5}$, $|\mathbf{b}|=\sqrt{10}$; $\cos\angle AOB = \dfrac{5}{\sqrt{50}} = \dfrac{1}{\sqrt{2}}$; $\angle AOB = 45°$
5. $|\mathbf{u}|=5$; $\hat{\mathbf{u}} = \tfrac{3}{5}\mathbf{i}-\tfrac{4}{5}\mathbf{j}$; perp unit vectors $= \pm\left(\tfrac{4}{5}\mathbf{i}+\tfrac{3}{5}\mathbf{j}\right)$
Q1 (2 marks): $\mathbf{a}\cdot\mathbf{b} = 2-4-2=-4$; $|\mathbf{a}|=3$, $|\mathbf{b}|=3$; $\cos\theta = \tfrac{-4}{9}$; $\theta = \cos^{-1}(-\tfrac{4}{9}) \approx \mathbf{116°}$ [2].
Q2 (3 marks): (a) $\mathbf{p}\cdot\mathbf{q} = 12-12=0$ [1]. (b) $\mathbf{p}-\mathbf{q} = -\mathbf{i}+7\mathbf{j}$; $|\mathbf{p}-\mathbf{q}|=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$; unit vector $= \dfrac{-\mathbf{i}+7\mathbf{j}}{5\sqrt{2}}$ [2].
Q3 (3 marks): $\cos\angle AOB = \dfrac{6}{12} = \dfrac{1}{2}$; $\angle AOB = 60°$ [1]. $|AB|^2 = |\mathbf{b}-\mathbf{a}|^2 = |\mathbf{b}|^2 - 2\mathbf{a}\cdot\mathbf{b}+|\mathbf{a}|^2 = 16-12+9=13$; $|AB| = \sqrt{13}$ [2].
Five timed multi-technique questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering mixed vector questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.