2014 H2 Mathematics Paper 1 Question 9

Vectors II: Lines and Planes

Answers

{- x + 2 y + z = 0}

{m: \mathbf{r} = \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}, \mu \in \mathbb{R}}

{\left( \frac{18}{7}, \frac{15}{7}, - \frac{12}{7} \right)}

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

Since

{p}

and

{q}

are perpendicular, the normal vector of

{p}

is a direction vector of

{q}

\begin{align*} \mathbf{n_q'} &= \mathbf{d_l} \times \mathbf{n_p} \\ &= \begin{pmatrix} 2 \\ - 1 \\ 4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ - 3 \end{pmatrix} \\ &= \begin{pmatrix} - 5 \\ 10 \\ 5 \end{pmatrix} \\ &= 5 \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}

\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} &= \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \\ &= 0 \end{align*}

Cartesian equation of

{q:}

{- x + 2 y + z = 0 \; \blacksquare}

\begin{align} \qquad &p: x + 2 y - 3 z = 12 \\ \qquad &q: - x + 2 y + z = 0 \end{align}

Using a GC, equation for

{m:}

\mathbf{r} = \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}, \mu \in \mathbb{R} \; \blacksquare

{\overrightarrow{OB} = \begin{pmatrix}6+4\mu\\3+\mu\\2\mu\end{pmatrix}}

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix}6+4\mu\\3+\mu\\2\mu\end{pmatrix} - \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix}5+4\mu\\4+\mu\\-3+2\mu\end{pmatrix} \end{align*}

\begin{align*} & \left|\overrightarrow{AB}\right|^2 \\ = & (5+4\mu)^2+(4+\mu)^2+(-3+2\mu)^2 \\ =& (25 + 40 \mu + 16 \mu^2) + (16 + 8 \mu + \mu^2) + (9 - 12 \mu + 4 \mu^2) \\ =& 50 + 36 \mu + 21 \mu^2 \; \blacksquare \end{align*}

Let

{y = 50 + 36 \mu + 21 \mu^2}

When point on

{m}

is nearest to

{A}

{y}

is a minimum
Hence

\frac{\mathrm{d}y}{\mathrm{d}\mu} = 0

\frac{\mathrm{d}y}{\mathrm{d}\mu} = 42\mu + 36

When

\frac{\mathrm{d}y}{\mathrm{d}\mu} = 0,

\mu = -\frac{6}{7}

\frac{\mathrm{d}^2y}{\mathrm{d}\mu^2} = 42 > 0

{\Rightarrow y}

is a minimum
When

{\mu = -\frac{6}{7},}

{\overrightarrow{OB} = \begin{pmatrix}6+4(-\frac{6}{7})\\3-\frac{6}{7}\\2(-\frac{6}{7})\end{pmatrix}}

{\overrightarrow{OB} = \begin{pmatrix}\frac{18}{7}\\\frac{15}{7}\\- \frac{12}{7}\end{pmatrix}}

Coordinates of point on

{m}

nearest to

{A}

\left( \frac{18}{7}, \frac{15}{7}, - \frac{12}{7} \right)\; \blacksquare