2023 H2 Maths Paper 1 Q3 — Worked Solution

Part (a)

3 marks

Approach

We let $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ and use the fact that $\mathbf{c}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ , together with the given perpendicularity condition written as a dot product equal to zero.

Since $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ :

\mathbf{c} \cdot \mathbf{a} = 0,\quad\mathbf{c} \cdot \mathbf{b} = 0

Since $\left( \mathbf{a} \times \mathbf{b} + \mathbf{a} \right)$ is perpendicular to $\left( \mathbf{a} \times \mathbf{b} + \mathbf{b} \right)$ :

\left( \mathbf{c} + \mathbf{a} \right) \cdot \left( \mathbf{c} + \mathbf{b} \right) = 0

\Rightarrow \mathbf{c} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{a} \cdot \mathbf{b} = 0

Substituting $\mathbf{c} \cdot \mathbf{b} = 0$ and $\mathbf{a} \cdot \mathbf{c} = 0$ :

\Rightarrow \left| \mathbf{c} \right|^{2} + \mathbf{a} \cdot \mathbf{b} = 0

Rearranging, and using $\mathbf{a} \cdot \mathbf{b} = - 1$ :

\left| \mathbf{a} \times \mathbf{b} \right|^{2} = - \mathbf{a} \cdot \mathbf{b}

= - ( - 1)

= 1

To check your answer:

\left| \mathbf{a} \times \mathbf{b} \right|^{2} = 1 \geq 0

, consistent with a magnitude squared, and the derivation used only

\mathbf{a} \times \mathbf{b}\bot\mathbf{a}

and

\mathbf{a} \times \mathbf{b}\bot\mathbf{b}

, which hold for any vectors

\mathbf{a},\mathbf{b}

✓

Since $\left| \mathbf{a} \times \mathbf{b} \right|^{2} = 1$ and $\left| \mathbf{a} \times \mathbf{b} \right| \geq 0$ :

$\therefore$ $\left| \mathbf{a} \times \mathbf{b} \right| = 1$ (shown).

Part (b)

3 marks

Let $\theta$ be the angle between the directions of $\mathbf{a}$ and $\mathbf{b}$ , so $0^{\circ} \leq \theta \leq 180^{\circ}$ . Using the standard results $\mathbf{a} \cdot \mathbf{b} = \left| \mathbf{a} \right|\left| \mathbf{b} \right|\cos\theta$ and $\left| \mathbf{a} \times \mathbf{b} \right| = \left| \mathbf{a} \right|\left| \mathbf{b} \right|\sin\theta$ , dividing eliminates the factor $\left| \mathbf{a} \right|\left| \mathbf{b} \right|$ , so the magnitudes of $\mathbf{a}$ and $\mathbf{b}$ are never needed:

\tan\theta = \frac{\left| \mathbf{a} \times \mathbf{b} \right|}{\mathbf{a} \cdot \mathbf{b}}

= \frac{1}{- 1}

= - 1

To check your answer: since

0^{\circ} \leq \theta \leq 180^{\circ}

\sin\theta \geq 0

always; here

\mathbf{a} \cdot \mathbf{b} < 0

\cos\theta < 0

, meaning

\theta

is obtuse — consistent with the second-quadrant solution below ✓

Since $\tan\theta = - 1$ with $\theta$ obtuse:

$\therefore$ $\theta = 135^{\circ}$ .

2023 H2 Maths Paper 1, Question 3

Part (a)

Part (b)

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