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 $\sqrt 6$ .

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]

This question was answered fairly well by most candidates using a diversity of approaches.