Question
The diagram shows a parallelogram ABCD.
The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .
(i) Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{AD}}} \) .
(iii) Hence show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\) .
Find the coordinates of point C.
(i) Find \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} \).
(ii) Hence find angle A.
Hence, or otherwise, find the area of the parallelogram.
Answer/Explanation
Markscheme
(i) evidence of approach M1
e.g. \({\text{B}} – {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\) AG N0
(ii) evidence of approach (M1)
e.g. \({\text{D}} – {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) , \(\left( {\begin{array}{*{20}{c}}
2\\
5\\
5
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{AD}}}= \left( {\begin{array}{*{20}{c}}
1\\
3\\
2
\end{array}} \right)\) A1 N2
(iii) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{AD}}} \)
correct substitution A1
e.g. \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
3\\
2
\end{array}} \right)\)
\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\) AG N0
[5 marks]
evidence of combining vectors (there are at least 5 ways) (M1)
e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AC}}} \) , \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{AD}}} \), \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OD}}} \)
correct substitution A1
\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}}
7\\
7\\
6
\end{array}} \right)} \right)\)
e.g. coordinates of C are \((7{\text{, }}7{\text{, }}6)\) A1 N1
[3 marks]
(i) evidence of using scalar product on \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) (M1)
e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 5(1) + 2(3) + 1(2)\)
\(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 13\) A1 N2
(ii) \(\left| {\overrightarrow {{\rm{AB}}} } \right| = 5.477 \ldots \) , \(\left| {\overrightarrow {{\rm{AD}}} } \right| = 3.741 \ldots \) (A1)(A1)
evidence of using \(\cos A = \frac{{\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} }}{{\left| {\overrightarrow {{\rm{AB}}} } \right|\left| {\overrightarrow {{\rm{AD}}} } \right|}}\) (M1)
correct substitution A1
e.g. \(\cos A = \frac{{13}}{{20.493}}\)
\(\widehat A = 0.884\) \((50.6^\circ )\) A1 N3
[7 marks]
METHOD 1
evidence of using \({\rm{area}} = 2\left( {\frac{1}{2}\left| {\overrightarrow {{\rm{AD}}} } \right|\left| {\overrightarrow {{\rm{AB}}} } \right|\sin {\rm{D}}\widehat {\rm{A}}{\rm{B}}} \right)\) (M1)
correct substitution A1
e.g. \({\rm{area}} = 2\left( {\frac{1}{2}(3,741 \ldots )(5.477 \ldots )\sin 0.883 \ldots } \right)\)
\({\rm{area}} = 15.8\) A1 N2
METHOD 2
evidence of using \({\rm{area}} = b \times h\) (M1)
finding height of parallelogram A1
e.g. \(h = 3.741 \ldots \times \sin 0.883 \ldots ( = 2.892 \ldots )\) , \(h = 5.477 \ldots \times \sin 0.883 \ldots ( = 4.234 \ldots )\)
\({\rm{area}} = 15.8\) A1 N2
[3 marks]
Question
The following diagram shows two perpendicular vectors u and v.
Let \(w = u – v\). Represent \(w\) on the diagram above.
Given that \(u = \left( \begin{array}{c}3\\2\\1\end{array} \right)\) and \(v = \left( \begin{array}{c}5\\n\\3\end{array} \right)\), where \(n \in \mathbb{Z}\), find \(n\).
Markscheme
METHOD 1 A1A1 N2
Note: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).
METHOD 2
A1A1 N2
Notes: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).
Additional lines not required.
[2 marks]
evidence of setting scalar product equal to zero (seen anywhere) R1
eg u \( \cdot \) v \( = 0,{\text{ }}15 + 2n + 3 = 0\)
correct expression for scalar product (A1)
eg \(3 \times 5 + 2 \times n + 1 \times 3,{\text{ }}2n + 18 = 0\)
attempt to solve equation (M1)
eg \(2n = – 18\)
\(n = – 9\) A1 N3
[4 marks]
Examiners report
Question
Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.
Let \({\mathop {{\text{PR}}}\limits^ \to }\) = 6i − j + 3k.
Find \(\mathop {{\text{PQ}}}\limits^ \to \).
Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|\).
Find the angle between PQ and PR.
Find the area of triangle PQR.
Hence or otherwise find the shortest distance from R to the line through P and Q.
Answer/Explanation
Markscheme
valid approach (M1)
eg (7, 4, 9) − (3, 2, 5) A − B
\(\mathop {{\text{PQ}}}\limits^ \to = \) 4i + 2j + 4k \(\left( { = \left( \begin{gathered}
4 \hfill \\
2 \hfill \\
4 \hfill \\
\end{gathered} \right)} \right)\) A1 N2
[2 marks]
correct substitution into magnitude formula (A1)
eg \(\sqrt {{4^2} + {2^2} + {4^2}} \)
\(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6\) A1 N2
[2 marks]
finding scalar product and magnitudes (A1)(A1)
scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)
magnitude of PR = \(\sqrt {36 + 1 + 9} = \left( {6.782} \right)\)
correct substitution of their values to find cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) M1
eg cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 – 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355\)
0.581746
\({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 0.582 radians or \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 33.3° A1 N3
[4 marks]
correct substitution (A1)
eg \(\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582\)
area is 11.2 (sq. units) A1 N2
[2 marks]
recognizing shortest distance is perpendicular distance from R to line through P and Q (M1)
eg sketch, height of triangle with base \(\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}\)
correct working (A1)
eg \(\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ \)
3.72677
distance = 3.73 (units) A1 N2
[3 marks]