(i)     Find the lengths of the sides of the triangle.

(ii)     Find $\cos {\rm{B\hat AC}}$.

(i)     Show that $\overrightarrow {{\text{BC}}}  \times \overrightarrow {{\text{CA}}}  =$ −7i − 3j − 16k.

(ii)     Hence, show that the area of the triangle ABC is $\frac{1}{2}\sqrt {314}$.

Find the Cartesian equation of the plane containing the triangle ABC.

Find a vector equation of (AB).

The point D on (AB) is such that $\overrightarrow {{\text{OD}}}$ is perpendicular to $\overrightarrow {{\text{BC}}}$ where O is the origin.

(i)     Find the coordinates of D.

(ii)     Show that D does not lie between A and B.

(i)     $\overrightarrow {{\text{AB}}}  = \overrightarrow {{\text{OB}}}  - \overrightarrow {{\text{OA}}}  =$ 5i – j – 2k (or in column vector form)     (A1)

Note: Award A1 if any one of the vectors, or its negative, representing the sides of the triangle is seen.

$\overrightarrow {{\text{AB}}}  =$ |5i – j – 2k|= $\sqrt {30}$

$\overrightarrow {{\text{BC}}}  =$ |–i – 3j + k|= $\sqrt {11}$

$\overrightarrow {{\text{CA}}}  =$ |–4i + 4j + k|= $\sqrt {33}$     A2

Note: Award A1 for two correct and A0 for one correct.

(ii)     METHOD 1

$\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}$     M1A1

Note: Award M1 for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, A1 for the correct computation using their vectors.

$= \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)$     A1

Note: Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.

METHOD 2

using the cosine rule

$\cos {\text{BAC}} = \frac{{30 + 33 - 11}}{{2\sqrt {30} \sqrt {33} }}$     M1A1

$= \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)$     A1

[6 marks]

$\overrightarrow {{\text{BC}}}  \times \overrightarrow {{\text{CA}}}  = \left| {\begin{array}{*{20}{c}}   i&j&k \\   { - 1}&{ - 3}&1 \\   { - 4}&4&1 \end{array}} \right|$     A1

$= \left( {( - 3) \times 1 - 1 \times 4} \right)$i + $\left( {1 \times ( - 4) - ( - 1) \times 1} \right)$j + $\left( {( - 1) \times 4 - ( - 3) \times ( - 4)} \right)$k     M1A1

= –7i – 3j – 16k     AG

(ii)     the area of $\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}}  \times \overrightarrow {{\text{CA}}} } \right|$     (M1)

$\frac{1}{2}\sqrt {{{( - 7)}^2} + {{( - 3)}^2} + {{( - 16)}^2}}$     A1

$= \frac{1}{2}\sqrt {314}$     AG

[5 marks]

attempt at the use of “(r – a)$\cdot$n = 0”     (M1)

using r = xi + yj + zk, a = $\overrightarrow {{\text{OA}}}$ and n = –7i – 3j – 16k     (A1)

$7x + 3y + 16z = 47$     A1

Note: Candidates who adopt a 2-parameter approach should be awarded, A1 for correct 2-parameter equations for x, y and z; M1 for a serious attempt at elimination of the parameters; A1 for the final Cartesian equation.

[3 marks]

r = $\overrightarrow {{\text{OA}}}  + t\overrightarrow {{\text{AB}}}$ (or equivalent)     M1

r = (–i + 2j + 3k) + t (5i – j – 2k)     A1

Note: Award M1A0 if “r =” is missing.

Note: Accept forms of the equation starting with B or with the direction reversed.

[2 marks]

(i)     $\overrightarrow {{\text{OD}}}  =$ (–i + 2j + 3k) + t(5i – j – 2k)

statement that $\overrightarrow {{\text{OD}}}  \cdot \overrightarrow {{\text{BC}}}  = 0$     (M1)

$\left( {\begin{array}{*{20}{c}}   { - 1 + 5t} \\   {2 - t} \\   {3 - 2t} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   { - 3} \\   1 \end{array}} \right) = 0$     A1

$- 2 - 4t = 0{\text{ or }}t = - \frac{1}{2}$     A1

coordinates of D are $\left( { - \frac{7}{2},\frac{5}{2},4} \right)$     A1

Note: Different forms of $\overrightarrow {{\text{OD}}}$ give different values of t, but the same final answer.

(ii)     $t < 0 \Rightarrow$ D is not between A and B     R1

[5 marks]

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).

Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).