Specimen 2017 H2 Mathematics Paper 1 Question 6

Vectors I: Basics, Dot and Cross Products

Answers

(a)

{\mathbf{a} = k (\mathbf{b} + \mathbf{c}).}

(b)

\mathbf{v} = \begin{pmatrix} \frac{2}{3} \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} - \frac{1}{3} \\ 0 \\ 1 \end{pmatrix}.

The set represents a line passing through the point with position vector ${\begin{pmatrix} \frac{2}{3} \\ 0 \\ 0 \end{pmatrix}}$ and parallel to the direction vector ${\begin{pmatrix} - \frac{1}{3} \\ 0 \\ 1 \end{pmatrix}.}$

Full solutions

(a)

\begin{gather*} \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} \\ \mathbf{a} \times \mathbf{b} - \mathbf{c} \times \mathbf{a} = \mathbf{0} \\ \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} = \mathbf{0} \\ \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{0} \\ \end{gather*}

Hence

{\mathbf{a}}

is parallel to

{\mathbf{b} + \mathbf{c}}

\mathbf{a} = k (\mathbf{b} + \mathbf{c}) \; \blacksquare

where

{k \in \mathbb{R}}

(b)

Let

{\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}.}

\begin{align*} \begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ - 3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \\ \begin{pmatrix} - 3 y \\ z + 3 x \\ - y \end{pmatrix} &= \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \\ \end{align*}

\begin{align} && \quad -3y &= 0 \\ && \quad z+3x &= 2 \\ && \quad -y &= 0 \\ \end{align}

Solving

{(1), (2)}

and

{(3)}

with a GC, set of vectors

{\mathbf{v}:}

\left\{ \mathbf{v}: \mathbf{v} = \begin{pmatrix} \frac{2}{3} \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} - \frac{1}{3} \\ 0 \\ 1 \end{pmatrix}, \; \lambda \in \mathbb{R} \right\} \; \blacksquare

This set represents a line passing through the point with position vector

{\begin{pmatrix} \frac{2}{3} \\ 0 \\ 0 \end{pmatrix}}

and parallel to the direction vector

{\begin{pmatrix} - \frac{1}{3} \\ 0 \\ 1 \end{pmatrix}. \; \blacksquare}