Find \(\overrightarrow {{\rm{BC}}} \) .

Find a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .

Show that \(\overrightarrow {{\rm{AB}}} \) is perpendicular to \(\overrightarrow {{\rm{AC}}} \) .

evidence of appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ - 2}\\
{ - 3}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
3
\end{array}} \right)\)

\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
{ - 8}\\
{ - 1}\\
{ - 1}
\end{array}} \right)\)     A1     N2

[2 marks]

attempt to find the length of \(\overrightarrow {{\rm{AB}}} \)     (M1)

\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( - 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\)     (A1)

unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
3
\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}
{\frac{6}{7}}\\
{ - \frac{2}{7}}\\
{\frac{3}{7}}
\end{array}} \right)} \right)\)     A1     N2

[3 marks]

recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular     (M1)

correct substitution in a scalar product formula     A1

e.g. \((6) \times ( - 2) + ( - 2) \times ( - 3) + (3) \times (2)\) , \(\cos \theta  = \frac{{ - 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)

correct calculation     A1

e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)

therefore, they are perpendicular     AG     N0

[3 marks]

Part (a) was generally done well with candidates employing different correct methods to find the vector \(\overrightarrow {{\rm{BC}}} \) . Some candidates subtracted the given vectors in the wrong order and others simply added them. Calculation errors were seen with some frequency.

Many candidates did not appear to know how to find a unit vector in part (b). Some tried to write down the vector equation of a line, indicating no familiarity with the concept of unit vectors while others gave the vector (1, 1, 1) or wrote the same vector \(\overrightarrow {{\rm{AB}}} \) as a linear combination of i, j and k. A number of candidates correctly found the magnitude but did not continue on to write the unit vector.

Candidates were generally successful in showing that the vectors in part (c) were perpendicular. Many used the efficient approach of showing that the scalar product equalled zero, while others worked a little harder than necessary and used the cosine rule to find the angle between the two vectors.