2007 H2 Mathematics Paper 1 Question 8

Vectors II: Lines and Planes

Answers

{\left( \frac{5}{2}, \frac{3}{2}, \frac{11}{2} \right)}

{78.8^\circ}

{\frac{4}{7} \sqrt{14} \textrm{ units}}

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 3 \\ 1 \\ - 3 \end{pmatrix} \\ \end{align*}

Equation of line

l: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} - 3 \\ 1 \\ - 3 \end{pmatrix}, \lambda \in \mathbb{R}

Equation of plane

p: \mathbf{r} \cdot \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix} = 17

Let

{X}

denote the point of intersection of

{l}

and

{p,}

Substituting the equation of

{l}

into the equation of

{p,}

\begin{pmatrix} 1 - 3 \lambda \\ 2 + \lambda \\ 4 - 3 \lambda \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix} = 17

3 (1 - 3 \lambda) - (2 + \lambda) + 2 (4 - 3 \lambda) = 17

\begin{align*} 9 - 16 \lambda &= 17 \\ - 16 \lambda &= 8 \\ \lambda &= - \frac{1}{2} \end{align*}

\begin{align*} \overrightarrow{OX} &= \begin{pmatrix} 1 - 3 (- \frac{1}{2}) \\ 2 + (- \frac{1}{2}) \\ 4 - 3 (- \frac{1}{2}) \end{pmatrix} \\ &= \begin{pmatrix} \frac{5}{2} \\ \frac{3}{2} \\ \frac{11}{2} \end{pmatrix} \end{align*}

Coordinates of point of intersection:

X \left( \frac{5}{2}, \frac{3}{2}, \frac{11}{2} \right) \; \blacksquare

Let

{\theta}

denote the angle between

{l}

and

{p}

\left|\mathbf{d} \cdot \mathbf{n}\right| = \left| \mathbf{d} \right| \left| \mathbf{n} \right| \sin \theta

\begin{align*} \left|\begin{pmatrix} - 3 \\ 1 \\ - 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix} \right| &= \left| \begin{pmatrix} - 3 \\ 1 \\ - 3 \end{pmatrix} \right| \left| \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix} \right| \sin \theta \\ \left|- 16 \right| &= (\sqrt{19}) (\sqrt{14}) \sin \theta \end{align*}

\begin{align*} \sin \theta &= \frac{16}{(\sqrt{19})(\sqrt{14})} \\ \theta &= 78.8^\circ \; \blacksquare \end{align*}

Let

{C \left( 0, - 17, 0 \right)}

be a point on

{p}

\begin{align*} \overrightarrow{CA} &= \overrightarrow{OA} - \overrightarrow{OC} \\ &= \begin{pmatrix} 1 \\ 19 \\ 4 \end{pmatrix} \\ \end{align*}

Perpendicular distance from

{A}

{p}

\begin{align*} & = \left| \overrightarrow{CA} \cdot \mathbf{\hat{n}} \right| \\ &= \frac{\left|\begin{pmatrix} 1 \\ 19 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix} \right|}{\left| \begin{pmatrix} 3 \\ - 1 \\ 2 \end{pmatrix}\right|} \\ &= \frac{\left|3 - 19 + 8 \right|}{\sqrt{9 + 1 + 4}} \\ &= \frac{8}{\sqrt{14}} \\ &= \frac{4}{7} \sqrt{14} \textrm{ units} \; \blacksquare \end{align*}