2010 H2 Mathematics Paper 1 Question 10

Vectors II: Lines and Planes

Answers

\begin{align*} \mathbf{d} &= \begin{pmatrix} - 3 \\ 6 \\ 9 \end{pmatrix} \\ &= 3 \begin{pmatrix} - 1 \\ 2 \\ 3 \end{pmatrix} \\ &= 3 \mathbf{n} \end{align*}

Hence the direction vector of

{l}

is parallel to the normal vector of

{p}

Hence

{l}

is perpendicular to

{p}

{\left( \frac{17}{2}, 2, \frac{3}{2} \right)}

{\left( 19, - 19, - 30 \right)}

{348 \textrm{ units}^2}

l: \mathbf{r} = \begin{pmatrix} 10 \\ - 1 \\ - 3 \end{pmatrix} + \lambda \begin{pmatrix} - 3 \\ 6 \\ 9 \end{pmatrix}, \lambda \in \mathbb{R}

p: \mathbf{r} \cdot \begin{pmatrix} 1 \\ - 2 \\ - 3 \end{pmatrix} = 0

\begin{align*} \mathbf{d} &= \begin{pmatrix} - 3 \\ 6 \\ 9 \end{pmatrix} \\ &= 3 \begin{pmatrix} - 1 \\ 2 \\ 3 \end{pmatrix} \\ &= 3 \mathbf{n} \end{align*}

Hence the direction vector of

{l}

is parallel to the normal vector of

{p}

Hence

{l}

is perpendicular to

{p}

{\; \blacksquare}

Let

{X}

be the point of intersection of

{l}

and

{p}

Substituting equation of

{l}

into equation of

{p,}

\begin{pmatrix} 10 - 3 \lambda \\ - 1 + 6 \lambda \\ - 3 + 9 \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ - 2 \\ - 3 \end{pmatrix} = 0

10 - 3 \lambda - 2 (- 1 + 6 \lambda) - 3 (- 3 + 9 \lambda) = 0

\begin{align*} 21 - 42 \lambda &= 0 \\ - 42 \lambda &= - 21 \\ \lambda &= \frac{1}{2} \\ \end{align*}

\begin{align*} \overrightarrow{OX} &= \begin{pmatrix} 10 - 3 (\frac{1}{2}) \\ - 1 + 6 (\frac{1}{2}) \\ - 3 + 9 (\frac{1}{2}) \end{pmatrix} \\ &= \begin{pmatrix} \frac{17}{2} \\ 2 \\ \frac{3}{2} \end{pmatrix} \end{align*}

Coordinates of point of intersection:

{\left( \frac{17}{2}, 2, \frac{3}{2} \right) \; \blacksquare}

Substituting

{\overrightarrow{OA}}

into equation of

{l}

\begin{pmatrix} - 2 \\ 23 \\ 33 \end{pmatrix} = \begin{pmatrix} 10 - 3 \lambda \\ - 1 + 6 \lambda \\ - 3 + 9 \lambda \end{pmatrix}

\begin{align} \qquad 10 - 3 \lambda &= - 2 \\ \qquad - 1 + 6 \lambda &= 23 \\ \qquad - 3 + 9 \lambda &= 33 \end{align}

Since

{\lambda = 4}

is a unique solution to the above system of linear equations,

{A}

lies on

{l \; \blacksquare}

By ratio theorem,

\overrightarrow{OX} = \frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}

{\overrightarrow{OB} = 2\overrightarrow{OX}-\overrightarrow{OA}}

{\overrightarrow{OB} = 2\begin{pmatrix} \frac{17}{2} \\ 2 \\ \frac{3}{2} \end{pmatrix} - 2\begin{pmatrix} - 2 \\ 23 \\ 33 \end{pmatrix}}

{\overrightarrow{OB} = \begin{pmatrix} 19 \\ - 19 \\ - 30 \end{pmatrix}}

Coordinates of

{B \left( 19, - 19, - 30 \right) \; \blacksquare}

\begin{align*} & \textrm{Area of} \triangle OAB \\ = & \frac{1}{2} \left| \overrightarrow{OA} \times \overrightarrow{OB} \right | \\ = & \frac{1}{2} \left| \begin{pmatrix} - 2 \\ 23 \\ 33 \end{pmatrix} \times \begin{pmatrix} 19 \\ - 19 \\ - 30 \end{pmatrix} \right | \\ = & \frac{1}{2} \left| \begin{pmatrix} - 63 \\ 567 \\ - 399 \end{pmatrix} \right | \\ = & \frac{1}{2} \sqrt{3969 + 321489 + 159201} \\ = & 348 \textrm{ units}^2 \textrm{ (nearest whole number)} \; \blacksquare \end{align*}