2023 H2 Mathematics Paper 1 Question 3

Vectors I: Basics, Dot and Cross Products

Answers

(a)

{{\left| \mathbf{a} \times \mathbf{b} \right| = 1.}}

(b)

{{\theta = 135 \degree.}}

Full solutions

(a)

We note that ${{\mathbf{a} \cdot \left( \mathbf{a} \times \mathbf{b} \right) = 0}}$ and ${{\mathbf{b} \cdot \left( \mathbf{a} \times \mathbf{b} \right) = 0}}$ since ${{\mathbf{a} \times \mathbf{b}}}$ is perpendicular to ${{\mathbf{a}}}$ and ${{\mathbf{b}}}$

Since ${{\mathbf{a} \times \mathbf{b} + \mathbf{a}}}$ is perpendicular to ${{\mathbf{a} \times \mathbf{b} + \mathbf{b},}}$

\begin{gather*} \left( \left( \mathbf{a} \times \mathbf{b} \right) + \mathbf{a} \right) \cdot \left( \left( \mathbf{a} \times \mathbf{b} \right) + \mathbf{b} \right) = 0 \\ \left| \mathbf{a} \times \mathbf{b} \right|^2 + \mathbf{a} \cdot \left( \mathbf{a} \times \mathbf{b} \right) + \mathbf{b} \cdot \left( \mathbf{a} \times \mathbf{b} \right) + \mathbf{a} \cdot \mathbf{b} = 0 \\ \left| \mathbf{a} \times \mathbf{b} \right|^2 + \mathbf{a} \cdot \mathbf{b} = 0 \\ \left| \mathbf{a} \times \mathbf{b} \right|^2 - 1 = 0 \\ \left| \mathbf{a} \times \mathbf{b} \right|^2 = 1 \end{gather*}

Since ${{\left| \mathbf{a} \times \mathbf{b} \right| \geq 0,}}$

{\left| \mathbf{a} \times \mathbf{b} \right| = 1 \; \blacksquare}

(b)

Since ${{\left| \mathbf{a} \times \mathbf{b} \right| = 1,}}$

\begin{equation} \left| \mathbf{a} \right| \left| \mathbf{b} \right| \sin \theta = 1 \end{equation}

Since ${{\mathbf{a} \cdot \mathbf{b} = -1,}}$

\begin{equation} \left| \mathbf{a} \right| \left| \mathbf{b} \right| \cos \theta = -1 \end{equation}

Taking ${{(1) \div (2),}}$

\begin{align*} \tan \theta &= -1 \\ \theta &= 135 \degree \; \blacksquare \end{align*}

Question Commentary

Hopefully most students were able to use the perpendicularity condition and start by using the fact that the dot product is zero.

The next steps are potentially tricky, where we need to be careful about expanding the dot product over the addition symbol and observing that ${{\mathbf{a} \times \mathbf{b} }}$ is perpendicular to both ${{\mathbf{a} }}$ and ${{\mathbf{b}.}}$