Worked Solutions
Vectors — Worked Solutions (HSC Maths Extension 2)
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Worked examples for HSC Maths Extension 2 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.
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.
With 3D vectors, decide early whether you need the dot product (angles, perpendicularity, projections) or a parametric form (lines and points), then commit to it.
Example 1 — Angle between two 3D vectors
Question
Two vectors are $\underset{\sim}{a} = 2\underset{\sim}{i} - \underset{\sim}{j} + 2\underset{\sim}{k}$ and $\underset{\sim}{b} = \underset{\sim}{i} + 2\underset{\sim}{j} + 2\underset{\sim}{k}$. Find the angle between them, to the nearest degree.
Solution
Use $\cos\theta = \dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{a}||\underset{\sim}{b}|}$.
Dot product: $\underset{\sim}{a}\cdot\underset{\sim}{b} = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4$.
Magnitudes: $|\underset{\sim}{a}| = \sqrt{4 + 1 + 4} = 3$ and $|\underset{\sim}{b}| = \sqrt{1 + 4 + 4} = 3$.
So $\cos\theta = \dfrac{4}{9}$, giving $\theta = \cos^{-1}\dfrac49 \approx 64^\circ$.
Don't forget the magnitudes in the denominator — a common slip is to read $\cos\theta$ straight off the dot product.
The dot product is the bridge between vectors and angles: $\underset{\sim}{a}\cdot\underset{\sim}{b} = |\underset{\sim}{a}||\underset{\sim}{b}|\cos\theta$. Rearranging for the angle, $\cos\theta = \dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{a}||\underset{\sim}{b}|}$.
First the dot product — multiply matching components and add: $(2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4$.
Then each magnitude, which is the square root of the sum of squared components: $|\underset{\sim}{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$, and similarly $|\underset{\sim}{b}| = \sqrt{1 + 4 + 4} = 3$.
Now substitute: $\cos\theta = \dfrac{4}{3 \times 3} = \dfrac49$, so $\theta = \cos^{-1}\left(\dfrac49\right) \approx 64^\circ$. The positive cosine tells us it's an acute angle, which is a nice sanity check.
Angle via $\cos\theta = \dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{a}||\underset{\sim}{b}|}$.
- $\underset{\sim}{a}\cdot\underset{\sim}{b} = 2 - 2 + 4 = 4$
- $|\underset{\sim}{a}| = \sqrt{4+1+4} = 3$
- $|\underset{\sim}{b}| = \sqrt{1+4+4} = 3$
- $\cos\theta = \tfrac49 \Rightarrow \theta \approx 64^\circ$
Where the marks go
- 1 mark: Correct dot product $\underset{\sim}{a}\cdot\underset{\sim}{b} = 4$
- 1 mark: Correct magnitudes $|\underset{\sim}{a}| = |\underset{\sim}{b}| = 3$
- 1 mark: Correct angle $\theta \approx 64^\circ$
Key idea
The angle between vectors comes from $\cos\theta = \dfrac{\underset{\sim}{a}\cdot\underset{\sim}{b}}{|\underset{\sim}{a}||\underset{\sim}{b}|}$; the dot product alone is not the cosine.
Example 2 — Vector equation of a line and a point on it
Question
A line passes through $A(1, 0, 2)$ and $B(3, 4, 6)$. Find a vector equation for the line, and determine whether the point $P(2, 2, 4)$ lies on it.
Solution
A line needs a point and a direction. Direction $\overrightarrow{AB} = B - A = (3-1,\ 4-0,\ 6-2) = (2, 4, 4)$.
Vector equation: $\underset{\sim}{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}$, $\lambda \in \mathbb{R}$.
For $P(2, 2, 4)$, equate components. From the first: $1 + 2\lambda = 2 \Rightarrow \lambda = \tfrac12$. Check the others: $0 + 4(\tfrac12) = 2$ ✓ and $2 + 4(\tfrac12) = 4$ ✓.
All three agree at $\lambda = \tfrac12$, so $P$ lies on the line. The test is simple: a point is on the line only if one value of $\lambda$ satisfies every component.
To describe a line in 3D we use a starting point plus a direction we can slide along. We have point $A$, and a natural direction is $\overrightarrow{AB} = B - A = (3-1,\ 4-0,\ 6-2) = (2, 4, 4)$.
So the vector equation is $\underset{\sim}{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}$, where $\lambda$ is a parameter that takes us to any point on the line.
To test $P(2, 2, 4)$, we ask: is there a single $\lambda$ that reaches it? Looking at the $x$-coordinate, $1 + 2\lambda = 2$ gives $\lambda = \tfrac12$. The key is that this same $\lambda$ must also work for $y$ and $z$: $y = 0 + 4(\tfrac12) = 2$ ✓ and $z = 2 + 4(\tfrac12) = 4$ ✓.
Because one value of $\lambda$ satisfies all three, the point really is on the line. If even one component had disagreed, $P$ would be off the line.
Point + direction; then test the point componentwise.
- $\overrightarrow{AB} = (2, 4, 4)$
- $\underset{\sim}{r} = (1, 0, 2) + \lambda(2, 4, 4)$
- $P$: $1 + 2\lambda = 2 \Rightarrow \lambda = \tfrac12$
- Check: $4(\tfrac12) = 2$ ✓, $2 + 4(\tfrac12) = 4$ ✓
- Consistent $\Rightarrow P$ lies on the line
Where the marks go
- 1 mark: Correct direction vector $\overrightarrow{AB} = (2, 4, 4)$
- 1 mark: Correct vector equation of the line
- 1 mark: Solves one component for $\lambda = \frac12$
- 1 mark: Verifies all components agree and concludes $P$ lies on the line
Key idea
A line is $\underset{\sim}{r} = \underset{\sim}{a} + \lambda\underset{\sim}{d}$; a point lies on it only if a single $\lambda$ satisfies every coordinate equation.