2021 H2 Mathematics Paper 1 Question 8

Vectors II: Lines and Planes

Answers

{x - 2 z = 3}

{\lambda = -1, \mu = 1}

{\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}, \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}}

{2 \sqrt{6} \textrm{ units}}

Let

{A (3,2,0)}

be a point on

{l_1}

and

{B = (1,-3,-1)}

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 2 \\ - 5 \\ - 1 \end{pmatrix} \end{align*}

\begin{align*} \mathbf{n'} &= \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} - 2 \\ - 5 \\ - 1 \end{pmatrix} \\ &= \begin{pmatrix} 8 \\ 0 \\ - 16 \end{pmatrix} \\ \mathbf{n} &= \begin{pmatrix} 1 \\ 0 \\ - 2 \end{pmatrix} \end{align*}

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

Cartesian equation of plane:

{x-2z=3 \; \blacksquare}

\begin{align*} \mathbf{d_1} \cdot \mathbf{d_2} &= \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \\ &= 2 - 6 + 4 \\ &= 0 \end{align*}

Hence the direction vectors of

{l_1}

and

{l_2}

are perpendicular.
Hence

{l_1}

is perpendicular to

{d_2}

{\blacksquare}

(ci)

\mathbf{r_1}-\mathbf{r_2}= \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix}

\begin{gather*} (\mathbf{r_1} - \mathbf{r_2}) \cdot \mathbf{d_1} = 0 \\ \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} = 0 \\ 14 \lambda = -14 \\ \lambda = -1 \; \blacksquare \end{gather*}

\begin{gather*} (\mathbf{r_1} - \mathbf{r_2}) \cdot \mathbf{d_2} = 0 \\ \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = 0 \\ - 21 \mu = -21 \\ \mu = 1 \; \blacksquare \end{gather*}

When

{\lambda = -1, \mathbf{r_1} = \begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}}

When

{\mu = 1, \mathbf{r_2} = \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}}

Hence the common perpendicular meets

{l_1}

and

{l_2}

at points with position vectors

{\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}}

and

{\begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix} \; \blacksquare}

(cii)

\begin{align*} & \textrm{Length of perpendicular } \\ &= \left|\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix} - \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix} \right| \\ &= \left|\begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix} \right| \\ &= 2 \sqrt{6} \textrm{ units} \; \blacksquare \end{align*}