Vectors — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 vectors. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Extension 1 — Vectors. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Projection of one vector onto another
Question
Let $\underset{\sim}{a} = 3\underset{\sim}{i} + 4\underset{\sim}{j}$ and $\underset{\sim}{b} = 5\underset{\sim}{i} + 12\underset{\sim}{j}$. Find the scalar projection of $\underset{\sim}{a}$ onto $\underset{\sim}{b}$.
Solution
Scalar projection of $\underset{\sim}{a}$ onto $\underset{\sim}{b}$ is $\dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{b}|}$.
Dot product: $\underset{\sim}{a}\cdot\underset{\sim}{b} = 3(5) + 4(12) = 15 + 48 = 63$.
Magnitude: $|\underset{\sim}{b}| = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$.
So the scalar projection $= \dfrac{63}{13}$.
Don't confuse scalar projection (divide by $|\underset{\sim}{b}|$ once) with vector projection (an extra factor of the unit vector).
The scalar projection tells you "how much of $\underset{\sim}{a}$ points along $\underset{\sim}{b}$", measured as a length. The formula is $\dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{b}|}$.
First the dot product — multiply matching components and add: $\underset{\sim}{a}\cdot\underset{\sim}{b} = (3)(5) + (4)(12) = 15 + 48 = 63$.
Next the length of $\underset{\sim}{b}$, since we project onto $\underset{\sim}{b}$: $|\underset{\sim}{b}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Dividing gives the scalar projection $= \dfrac{63}{13}$.
We divide by $|\underset{\sim}{b}|$ (not $|\underset{\sim}{a}|$) because the projection is measured along the direction of $\underset{\sim}{b}$.
Scalar projection $= \dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{b}|}$.
- $\underset{\sim}{a}\cdot\underset{\sim}{b} = 3(5)+4(12) = 63$
- $|\underset{\sim}{b}| = \sqrt{25+144} = 13$
- Projection $= \dfrac{63}{13}$
Where the marks go
- 1 mark: Correct dot product $\underset{\sim}{a}\cdot\underset{\sim}{b} = 63$
- 1 mark: Correct magnitude $|\underset{\sim}{b}| = 13$
- 1 mark: Correct scalar projection $\frac{63}{13}$
Key idea
Scalar projection of $\underset{\sim}{a}$ onto $\underset{\sim}{b}$ is $\frac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{b}|}$ — dot product over the length of the vector you project onto.
Example 2 — Vector geometry: diagonals of a rhombus
Question
$OACB$ is a rhombus with $\overrightarrow{OA} = \underset{\sim}{a}$ and $\overrightarrow{OB} = \underset{\sim}{b}$, where $|\underset{\sim}{a}| = |\underset{\sim}{b}|$. Using vectors, prove that the diagonals $OC$ and $AB$ are perpendicular.
Solution
Since $OACB$ is a rhombus, $\overrightarrow{OC} = \underset{\sim}{a} + \underset{\sim}{b}$ and $\overrightarrow{AB} = \underset{\sim}{b} - \underset{\sim}{a}$.
Test perpendicularity with the dot product:
$$\overrightarrow{OC}\cdot\overrightarrow{AB} = (\underset{\sim}{a}+\underset{\sim}{b})\cdot(\underset{\sim}{b}-\underset{\sim}{a}) = |\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2.$$
Because all sides are equal, $|\underset{\sim}{a}| = |\underset{\sim}{b}|$, so this is $0$.
A zero dot product (with non-zero vectors) means the diagonals are perpendicular. The whole proof hinges on the equal-sides condition giving $|\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2 = 0$.
Let's set up the two diagonals as vectors. Going from $O$ to $C$ takes us across both sides, so $\overrightarrow{OC} = \underset{\sim}{a} + \underset{\sim}{b}$. The other diagonal runs from $A$ to $B$: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \underset{\sim}{b} - \underset{\sim}{a}$.
Two vectors are perpendicular exactly when their dot product is zero, so let's compute it:
$$\overrightarrow{OC}\cdot\overrightarrow{AB} = (\underset{\sim}{a}+\underset{\sim}{b})\cdot(\underset{\sim}{b}-\underset{\sim}{a}).$$
Expanding carefully (the dot product distributes): $= \underset{\sim}{a}\cdot\underset{\sim}{b} - \underset{\sim}{a}\cdot\underset{\sim}{a} + \underset{\sim}{b}\cdot\underset{\sim}{b} - \underset{\sim}{b}\cdot\underset{\sim}{a}$. The mixed terms cancel, leaving $|\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2$.
Here's where being a rhombus matters: all sides are equal length, so $|\underset{\sim}{a}| = |\underset{\sim}{b}|$ and the result is $0$. A zero dot product means the diagonals meet at right angles. This is why the rhombus condition is essential — without equal sides the diagonals would not be perpendicular.
- $\overrightarrow{OC} = \underset{\sim}{a}+\underset{\sim}{b}$, $\overrightarrow{AB} = \underset{\sim}{b}-\underset{\sim}{a}$
- $\overrightarrow{OC}\cdot\overrightarrow{AB} = (\underset{\sim}{a}+\underset{\sim}{b})\cdot(\underset{\sim}{b}-\underset{\sim}{a}) = |\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2$
- Rhombus ⇒ $|\underset{\sim}{a}| = |\underset{\sim}{b}|$ ⇒ dot product $= 0$
- Zero dot product ⇒ $OC \perp AB$.
Where the marks go
- 1 mark: Expresses both diagonals as $\underset{\sim}{a}+\underset{\sim}{b}$ and $\underset{\sim}{b}-\underset{\sim}{a}$
- 1 mark: Forms the dot product of the two diagonals
- 1 mark: Simplifies to $|\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2$
- 1 mark: Uses $|\underset{\sim}{a}| = |\underset{\sim}{b}|$ to conclude perpendicularity
Key idea
Two vectors are perpendicular when their dot product is zero; $(\underset{\sim}{a}+\underset{\sim}{b})\cdot(\underset{\sim}{b}-\underset{\sim}{a}) = |\underset{\sim}{b}|^2 - |\underset{\sim}{a}|^2$, which vanishes when the magnitudes are equal.
Frequently asked questions
Step-by-step solutions to exam-style questions on Vectors in HSC Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 1 syllabus for Vectors, using the methods and notation expected in the exam.