| Date | May 2009 | Marks available | 6 | Reference code | 09M.1.hl.TZ1.8 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Show that | Question number | 8 | Adapted from | N/A |
Question
A triangle has vertices A(1, −1, 1), B(1, 1, 0) and C(−1, 1, −1) .
Show that the area of the triangle is \(\sqrt 6 \) .
Markscheme
METHOD 1
for finding two of the following three vectors (or their negatives)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
0 \\
2 \\
{ - 1}
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
{ - 2}
\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
0 \\
{ - 1}
\end{array}} \right)\) (A1)(A1)
and calculating
EITHER
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
0&2&{ - 1} \\
{ - 2}&2&{ - 2}
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
4
\end{array}} \right)\) M1A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) M1
OR
\(\overrightarrow {{\text{BA}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
0&{ - 2}&1 \\
{ - 2}&0&{ - 1}
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
2 \\
{ - 2} \\
{ - 4}
\end{array}} \right)\) M1A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BA}}} \times \overrightarrow {{\text{BC}}} } \right|\) M1
OR
\(\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
2&{ - 2}&2 \\
2&0&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
4
\end{array}} \right)\) M1A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|\) M1
THEN
area \(\Delta {\text{ABC}} = \frac{{\sqrt {24} }}{2}\) A1
\( = \sqrt 6 \) AG N0
METHOD 2
for finding two of the following three vectors (or their negatives)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
0 \\
2 \\
{ - 1}
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
{ - 2}
\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
0 \\
{ - 1}
\end{array}} \right)\) (A1)(A1)
EITHER
\(\cos A = \frac{{\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\) M1
\( = \frac{6}{{\sqrt 5 \sqrt {12} }} = \frac{6}{{\sqrt {60} }}{\text{ }}\left( {{\text{or }}\frac{3}{{\sqrt {15} }}} \right)\)
\(\sin A = \sqrt {\frac{2}{5}} \) A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|\sin A\) M1
\( = \frac{1}{2}\sqrt 5 \sqrt {12} \sqrt {\frac{2}{5}} \)
\( = \frac{1}{2}\sqrt {24} \) A1
\( = \sqrt 6 \) AG N0
OR
\(\cos B = \frac{{\overrightarrow {{\text{BA}}} \cdot \overrightarrow {{\text{BC}}} }}{{\left| {\overrightarrow {{\text{BA}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|}}\) M1
\( = - \frac{1}{{\sqrt 5 \sqrt 5 }} = - \frac{1}{5}\)
\(\sin B = \sqrt {\frac{{24}}{{25}}} {\text{ }}\left( {{\text{or }}\frac{{\sqrt {24} }}{5}} \right)\) A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BA}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\sin B\) M1
\( = \frac{1}{2}\sqrt 5 \sqrt 5 \sqrt {\frac{{24}}{{25}}} \)
\( = \frac{1}{2}\sqrt {24} \) A1
\( = \sqrt 6 \) AG N0
OR
\(\cos C = \frac{{\overrightarrow {{\text{CA}}} \cdot \overrightarrow {{\text{CB}}} }}{{\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CB}}} } \right|}}\) M1
\( = \frac{6}{{\sqrt {12} \sqrt 5 }} = \frac{6}{{\sqrt {60} }}{\text{ }}\left( {{\text{or }}\frac{3}{{\sqrt {15} }}} \right)\)
\(\sin C = \sqrt {\frac{2}{5}} \) A1
area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CB}}} } \right|\sin C\) M1
\( = \frac{1}{2}\sqrt {12} \sqrt 5 \sqrt {\frac{2}{5}} \)
\( = \frac{1}{2}\sqrt {24} \) A1
\( = \sqrt 6 \) AG N0
METHOD 3
for finding two of the following three vectors (or their negatives)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
0 \\
2 \\
{ - 1}
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
{ - 2}
\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
0 \\
{ - 1}
\end{array}} \right)\) (A1)(A1)
\({\text{AB}} = \sqrt 5 = c\) , \({\text{AC}} = \sqrt {12} = 2\sqrt 3 = b\) , \({\text{BC}} = \sqrt 5 = a\) M1A1
\(s = \frac{{\sqrt 5 + 2\sqrt 3 + \sqrt 5 }}{2} = \sqrt 3 + \sqrt 5 \) M1
area \(\Delta {\text{ABC}} = \sqrt {s(s - a)(s - b)(s - c)} \)
\( = \sqrt {(\sqrt 3 + \sqrt 5 )(\sqrt 3 )(\sqrt 5 - \sqrt 3 )(\sqrt 3 )} \)
\( = \sqrt {3(5 - 3)} \) A1
\( = \sqrt 6 \) AG N0
METHOD 4
for finding two of the following three vectors (or their negatives)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
0 \\
2 \\
{ - 1}
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
2 \\
{ - 2}
\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}
{ - 2} \\
0 \\
{ - 1}
\end{array}} \right)\) (A1)(A1)
\({\text{AB}} = {\text{BC}} = \sqrt 5 \) and \({\text{AC}} = \sqrt {12} = 2\sqrt 3 \) M1A1
\(\Delta {\text{ABC}}\) is isosceles
let M be the midpoint of [AC] , the height \({\text{BM}} = \sqrt {5 - 3} = \sqrt 2 \) M1
area \(\Delta {\text{ABC}} = \frac{{2\sqrt 3 \times \sqrt 2 }}{2}\) A1
\( = \sqrt 6 \) AG N0
[6 marks]
Examiners report
This question was answered fairly well by most candidates using a diversity of approaches.
