Worked Solutions
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.
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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.