2020 H2 Mathematics Paper 1 Question 5

Vectors I: Basics, Dot and Cross Products

Answers

{\mathbf{a}}

is parallel to

{\mathbf{b}}

The set of all possible positions of the point

{R}

forms a line passing through point

{P}

and parallel to the direction vector

{\overrightarrow{OQ}}

{3 x - 5 y + 2 z = - 5}

The set of all possible positions of the point

{R}

forms a plane passing through point

{P}

and perpendicular to the normal vector

{\overrightarrow{OQ}}

\begin{align*} \mathbf{a}\times\mathbf{b} &= \mathbf{b} \times \mathbf{a} \\ \mathbf{a}\times\mathbf{b} &= -\mathbf{a} \times \mathbf{b} \\ 2 \mathbf{a} \times \mathbf{b} &= 0 \\ \mathbf{a} \times \mathbf{b} &= 0 \end{align*}

Since

{\mathbf{a}}

and

{\mathbf{b}}

are non-zero and

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

{\mathbf{a}}

is parallel to

{\mathbf{b} \; \blacksquare}

(bi)

From (a),

{\mathbf{r}-\mathbf{p}}

is parallel to

{\mathbf{q}}

, or

{\mathbf{r}-\mathbf{p}=\mathbf{0}}

. Hence

\begin{gather*} \mathbf{r}-\mathbf{p} = \lambda \mathbf{q} \\ \mathbf{r} = \mathbf{p} + \lambda \mathbf{q}, \; \lambda \in \mathbb{R} \end{gather*}

Hence the set of all possible positions of the point

{R}

forms a line passing through point

{P}

and parallel to the direction vector

{\overrightarrow{OQ} \; \blacksquare}

(bii)

\begin{align*} (\mathbf{r}-\mathbf{p}) \cdot \mathbf{q} &= 0 \\ \mathbf{r} \cdot \mathbf{q} - \mathbf{p} \cdot \mathbf{q} &= 0 \\ \end{align*}

\begin{align*} \mathbf{r} \cdot \mathbf{q} &= \mathbf{p} \cdot \mathbf{q} \\ \mathbf{r} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} &= \begin{pmatrix} - 1 \\ 2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} \\ \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} &= - 5 \\ \end{align*}

3 x - 5 y + 2 z = - 5 \; \blacksquare

The set of all possible positions of the point

{R}

forms a plane passing through point

{P}

and perpendicular to the normal vector

{\overrightarrow{OQ} \; \blacksquare}