2014 H2 Mathematics Paper 1 Question 3

Vectors I: Basics, Dot and Cross Products

Answers

Either

{\mathbf{a}}

is parallel to

{\mathbf{b},}

{\mathbf{a} = \mathbf{0},}

{\mathbf{b} = \mathbf{0},}

{\mathbf{n} = \displaystyle \frac{1}{3} \begin{pmatrix} 1 \\ 2 \\ - 2 \end{pmatrix}}

{\cos \theta = \frac{2}{3}}

By the reasoning from part (i),

{\mathbf{n}}

is parallel to

{\mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k}}

Since

{\mathbf{n}}

is a unit vector, one possible answer is

\begin{align*} \mathbf{n} &= \frac{\mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k}}{\left|\mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k}\right|} \\ &= \frac{1}{3} \begin{pmatrix} 1 \\ 2 \\ - 2 \end{pmatrix} \; \blacksquare \end{align*}

\left| \mathbf{a} \cdot \mathbf{b} \right| = \left| \mathbf{a} \right| \left| \mathbf{b} \right| \cos \theta

Let

{\theta}

be the acute angle between

{\mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k}}

and the

{z}

-axis

\begin{align*} \left| \begin{pmatrix} 1 \\ 2 \\ - 2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right | &= \left|\begin{pmatrix} 1 \\ 2 \\ - 2 \end{pmatrix}\right| \left|\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right| \cos\theta \\ \cos \theta &= \frac{2}{3} \; \blacksquare \end{align*}