| Date | May 2008 | Marks available | 3 | Reference code | 08M.1.sl.TZ2.8 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 8 | Adapted from | N/A |
Question
Consider the points A (1 , 5 , 4) , B (3 , 1 , 2) and D (3 , k , 2) , with (AD) perpendicular to (AB) .
The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .
Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AD}}} \) giving your answer in terms of k .
[3 marks]
Show that \(k = 7\) .
The point C is such that \(\overrightarrow {{\rm{BC}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} \) .
Find the position vector of C.
Find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}\) .
Markscheme
(i) evidence of combining vectors (M1)
e.g. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}} \) (or \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) in part (ii))
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
2\\
{ - 4}\\
{ - 2}
\end{array}} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
2\\
{k - 5}\\
{ - 2}
\end{array}} \right)\) A1 N1
[3 marks]
evidence of using perpendicularity \( \Rightarrow \) scalar product = 0 (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
{ - 4}\\
{ - 2}
\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}
2\\
{k - 5}\\
{ - 2}
\end{array}} \right) = 0\)
\(4 - 4(k - 5) + 4 = 0\) A1
\( - 4k + 28 = 0\) (accept any correct equation clearly leading to \(k = 7\) ) A1
\(k = 7\) AG N0
[3 marks]
\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
2\\
2\\
{ - 2}
\end{array}} \right)\) (A1)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ - 1}
\end{array}} \right)\) A1
evidence of correct approach (M1)
e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} \) , \(\left( {\begin{array}{*{20}{c}}
3\\
1\\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ - 1}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{x - 3}\\
{y - 1}\\
{z - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ - 1}
\end{array}} \right)\)
\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2\\
1
\end{array}} \right)\) A1 N3
[4 marks]
METHOD 1
choosing appropriate vectors, \(\overrightarrow {{\rm{BA}}} \) , \(\overrightarrow {{\rm{BC}}} \) (A1)
finding the scalar product M1
e.g. \( - 2(1) + 4(1) + 2( - 1)\) , \(2(1) + ( - 4)(1) + ( - 2)( - 1)\)
\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) A1 N1
METHOD 2
\(\overrightarrow {{\rm{BC}}} \) parallel to \(\overrightarrow {{\rm{AD}}} \) (may show this on a diagram with points labelled) R1
\(\overrightarrow {{\rm{BC}}} \bot \overrightarrow {{\rm{AB}}} \) (may show this on a diagram with points labelled) R1
\({\rm{A}}\widehat {\rm{B}}{\rm{C}} = 90^\circ \)
\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) A1 N1
[3 marks]
Examiners report
This question was well done by many candidates. Most found \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) correctly.
The majority of candidates correctly used the scalar product to show \(k = 7\) .
Some confusion arose in substituting \(k = 7\) into \(\overrightarrow {{\rm{AD}}} \) , but otherwise part (c) was well done, though finding the position vector of C presented greater difficulty.
Owing to \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{BC}}} \) being perpendicular, no problems were created by using these two vectors to find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = 0\) , and the majority of candidates answering part (d) did exactly that.
