Show that \(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\) .

Let C and D be points such that ABCD is a rectangle.

Given that \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
4\\
p\\
1
\end{array}} \right)\) , show that \(p = 3\) .

Let C and D be points such that ABCD is a rectangle.

Find the coordinates of point C.

Let C and D be points such that ABCD is a rectangle.

Find the area of rectangle ABCD.

correct approach     A1

e.g. \(\overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}} ,\left( {\begin{array}{*{20}{c}}
6\\
0\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\)

\(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\)    AG     N0

[1 mark]

recognizing \(\overrightarrow {{\rm{AD}}} \) is perpendicular to \(\overrightarrow {{\rm{AB}}} \)  (may be seen in sketch)     (R1) 

e.g. adjacent sides of rectangle are perpendicular

recognizing dot product must be zero     (R1)

e.g. \(\overrightarrow {{\rm{AD}}}  \bullet \overrightarrow {AB}  = 0\)

correct substitution     (A1)

e.g. \((1 \times 4) + ( - 2 \times p) + (2 \times 1)\) , \(4 - 2p + 2 = 0\)

equation which clearly leads to \(p = 3\)     A1 

e.g. \(6 - 2p = 0\) , \(2p = 6\) 

\(p = 3\)     AG     N0

[4 marks]

correct approach (seen anywhere including sketch)     (A1)

e.g. \(\overrightarrow {{\rm{OC}}}  = \overrightarrow {{\rm{OB}}}  + \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{OD}}}  + \overrightarrow {{\rm{DC}}} \)

recognizing opposite sides are equal vectors (may be seen in sketch)     (R1)

e.g. \(\overrightarrow {{\rm{BC}}}  = \overrightarrow {{\rm{AD}}} \) , \(\overrightarrow {{\rm{DC}}}  = \overrightarrow {{\rm{AB}}} \) , \(\left( {\begin{array}{*{20}{c}}
6\\
0\\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
4\\
3\\
1
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
9\\
5\\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\)

coordinates of point C are (10, 3, 4) (accept \(\left( {\begin{array}{*{20}{c}}
{10}\\
3\\
4
\end{array}} \right)\) )    A2     N4

Note: Award A1 for two correct values.

[4 marks]

attempt to find one side of the rectangle     (M1)

e.g. substituting into magnitude formula

two correct magnitudes     A1A1

e.g. \(\sqrt {{{(1)}^2} + {{( - 2)}^2} + {2^2}} \) , 3 ; \(\sqrt {16 + 9 + 1} \) , \(\sqrt {26} \)

multiplying magnitudes     (M1)

e.g. \(\sqrt {26} \times \sqrt 9 \)

\({\rm{area}} = \sqrt {234} ( = 3\sqrt {26} )\) (accept \(3 \times \sqrt {26} \) )     A1     N3

[5 marks]

Part (a) was answered correctly by nearly every candidate.

In part (b), the candidates who realized that the vectors must be perpendicular were successful using the scalar product to find p . Incorrect approaches included using magnitudes, or creating vector equations of lines for the sides and setting them equal to each other. In addition, there were a good number of candidates who worked backwards, using the given value of 3 for p to find the coordinates of point D. Candidates who work backwards on a "show that" question will earn no marks.

Part (c) was more difficult for candidates, and was left blank by some. Some candidates found \(\overrightarrow {{\rm{AC}}} \) rather than \(\overrightarrow {{\rm{OC}}} \) , as required. Many candidates recognized that the opposite sides of the rectangle must be equal, but did not consider the directions of the vectors for those sides. There were also a good number of candidates who mislabelled the vertices of their rectangles, which led to them working with a rectangle ABDC, rather than ABCD.

The majority of candidates who attempted part (d) were successful in multiplying the magnitudes of the sides. Unfortunately, there were some who set up their solutions correctly, then had arithmetic errors in their working.