4400 Higher June 2006 Paper 4 Q14

EdexcelHigherOld spec8 marksVectors

14 \(OABC\) is a parallelogram.

Parallelogram OABC with diagonal OB drawn

Diagram NOT accurately drawn

\(\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\), \(\overrightarrow{OC} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}\).

(a) Find the vector \(\overrightarrow{OB}\) as a column vector. (1)

\(X\) is the point on \(OB\) such that \(OX = kOB\), where \(0 < k < 1\)

(b) Find, in terms of \(k\), the vectors

(i) \(\overrightarrow{OX}\),

(ii) \(\overrightarrow{AX}\),

(iii) \(\overrightarrow{XC}\). (3)

(d) Use your answer to part (c) to show that the diagonals of the parallelogram \(OABC\) bisect one another. (2)

Scheme	Marks
\(\begin{pmatrix} 5 \\ 2 \end{pmatrix}\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
(i) \(k\begin{pmatrix} 5 \\ 2 \end{pmatrix}\) oe	B1f
(ii) \(-\begin{pmatrix} 1 \\ 2 \end{pmatrix} + k\begin{pmatrix} 5 \\ 2 \end{pmatrix}\) oe	B1f
(iii) \(\begin{pmatrix} 4 \\ 0 \end{pmatrix} - k\begin{pmatrix} 5 \\ 2 \end{pmatrix}\) oe	B1f
	(3)

Notes

(i) or \(k\overrightarrow{OB}\) oe; (ii) or \(-\overrightarrow{OA} + k\overrightarrow{OB}\) oe; (iii) or \(-k\overrightarrow{OB} + \overrightarrow{OC}\) oe

Not \(x\) or 0.5 for \(k\). Allow without brackets or arrows

Mark scheme (c)

Scheme	Marks
\(-\begin{pmatrix} 1 \\ 2 \end{pmatrix} + k\begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} - k\begin{pmatrix} 5 \\ 2 \end{pmatrix}\) oe	M1
\(k = \frac{1}{2}\)	A1
	(2)

Notes

M1: or \(-1 + 5k = 4 - 5k\) or \(-2 + 2k = -2k\) \(\quad\) ft(b) for M1 only

A1: No wking, \(k = 0.5\): M1A1

Mark scheme (d)

Scheme	Marks
\(k = \frac{1}{2} \Rightarrow X\) is midpt of \(OB\)	B1
\(\overrightarrow{AX} = \overrightarrow{XC} \Rightarrow X\) is midpt of \(AC\)	B1
	(2)
	(8 marks)

Notes

No marks unless (c) 2 mks
“\(k = \frac{1}{2} \Rightarrow X\) is midpt of \(OB\) & \(AC\)” or “\(k = \frac{1}{2} \Rightarrow X\) is midpt of //m” : B1

B1: Allow without arrows