Explain why ABCD is a parallelogram.

Using vector algebra, show that $\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to$.

Show that p = 1, q = 1 and r = 4.

Find the area of the parallelogram ABCD.

Find the vector equation of the straight line passing through M and normal to the plane $\Pi$ containing ABCD.

Find the Cartesian equation of $\Pi$.

Find the coordinates of X, Y and Z.

Find YZ.

a pair of opposite sides have equal length and are parallel      R1

hence ABCD is a parallelogram      AG

[1 mark]

attempt to rewrite the given information in vector form       M1

b − a = c − d      A1

rearranging d − a = c − b       M1

hence  $\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to$     AG

Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.

[3 marks]

EITHER

use of $\mathop {{\text{AB}}}\limits^ \to   = \mathop {{\text{DC}}}\limits^ \to$     (M1)

$\left( \begin{gathered} 2 \hfill \\ - 3 \hfill \\ p + 3 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q + 1 \hfill \\ 1 - r \hfill \\ 4 \hfill \\ \end{gathered} \right)$       A1A1

OR

use of $\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to$      (M1)

$\left( \begin{gathered} - 2 \hfill \\ r - 2 \hfill \\ 1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q - 3 \hfill \\ 2 \hfill \\ 2 - p \hfill \\ \end{gathered} \right)$      A1A1

THEN

attempt to compare coefficients of i, j, and k in their equation or statement to that effect       M1

clear demonstration that the given values satisfy their equation       A1
p = 1, q = 1, r = 4       AG

[5 marks]

attempt at computing $\mathop {{\text{AB}}}\limits^ \to  \, \times \mathop {{\text{AD}}}\limits^ \to$ (or equivalent)       M1

$\left( \begin{gathered} - 11 \hfill \\ - 10 \hfill \\ - 2 \hfill \\ \end{gathered} \right)$     A1

area $= \left| {\mathop {{\text{AB}}}\limits^ \to  \, \times \mathop {{\text{AD}}}\limits^ \to  } \right|\left( { = \sqrt {225} } \right)$      (M1)

= 15       A1

[4 marks]

valid attempt to find $\mathop {{\text{OM}}}\limits^ \to   = \left( {\frac{1}{2}\left( {a + c} \right)} \right)$      (M1)

$\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ - \frac{1}{2} \hfill \\ \end{gathered} \right)$     A1

the equation is

r = $\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ - \frac{1}{2} \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} 11 \hfill \\ 10 \hfill \\ 2 \hfill \\ \end{gathered} \right)$ or equivalent       M1A1

Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.

[4 marks]

attempt to obtain the equation of the plane in the form ax + by + cz = d       M1

11x + 10y + 2z = 25      A1A1

Note: A1 for right hand side, A1 for left hand side.

[3 marks]

putting two coordinates equal to zero       (M1)

${\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)$      A1

[2 marks]

${\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}}$     M1

$= \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)$     A1

[4 marks]