October 2021 Paper 2 Q9

OCR ACurrent spec6 marksVectors

9 Points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) relative to an origin \(O\) in 3-dimensional space. Rectangles \(OADC\) and \(BEFG\) are the base and top surface of a cuboid.

Cuboid with base OADC and top BEFG; vectors a along OA, b along OB (vertical) and c along OC; M is the midpoint of BC and X lies on the dashed line from A to M

The point \(M\) is the midpoint of \(BC\).
The point \(X\) lies on \(AM\) such that \(AX = 2XM\).

(a) Find \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), simplifying your answer. [4]

(b) Hence show that the lines \(OF\) and \(AM\) intersect. [2]

Mark scheme (a)

Scheme	Marks
Summary method: \(\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})\) or \(\mathbf{b} + \frac{1}{2}(-\mathbf{b} + \mathbf{c})\) oe	B1
\(\overrightarrow{AM}\) or \(\overrightarrow{MA}\) attempted in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) \(\left(= \pm\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\ \text{oe}\right)\)	M1
\(\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}\overrightarrow{AM}\) or \(\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MA}\) oe attempted in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\)	M1
\(\overrightarrow{OX} = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\)	A1
	[4]

Notes

B1: Can be implied

M1: May be included in working, eg \(\overrightarrow{AX} = \frac{2}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)
Not necessarily correct

M1: Not necessarily correct

A1: or equivalent simplified form

Examples of methods using the above Other correct methods may be seen Allow inadequate notation

Scheme	Marks
\(\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})\)	B1
\(\overrightarrow{AX} = \frac{2}{3}\overrightarrow{AM} = \frac{2}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)	M1
\(\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}\overrightarrow{AM} = \mathbf{a} + \frac{2}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)	M1
\(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\)	A1

M1: for \(\overrightarrow{AM} = \left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\) implied

Scheme	Marks
\(\overrightarrow{BM} = \frac{1}{2}(-\mathbf{b} + \mathbf{c})\) \(\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{BM}\) \(= -\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c}\)	B1 M1
\(\overrightarrow{OX} = \mathbf{a} + \frac{2}{3}\left(-\mathbf{a} + \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c}\right)\)	M1
\(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\)	A1

B1: Implied

Scheme	Marks
\(\overrightarrow{OM} = \frac{1}{2}(\mathbf{b} + \mathbf{c})\)	B1
\(\overrightarrow{XM} = \frac{1}{3}\overrightarrow{AM}\) \(= \frac{1}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)	M1
\(\overrightarrow{OX} = \overrightarrow{OM} - \frac{1}{3}\overrightarrow{AM}\) \(= \frac{1}{2}(\mathbf{b} + \mathbf{c}) - \frac{1}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)	M1
\(= \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\)	A1

M1: for \(\overrightarrow{AM} = \left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\) implied
\(\overrightarrow{OX} = \overrightarrow{OM} - \frac{1}{3}\overrightarrow{AM}\) is equivalent to \(\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MA}\)

Mark scheme (b)

Scheme	Marks
\(\overrightarrow{OF} = \mathbf{a} + \mathbf{b} + \mathbf{c}\)	B1*
Hence \(X\) lies on \(OF\), so \(AM\) and \(OF\) intersect	B1dep
	[2]

Notes

B1dep: Both statements needed. NB dep on B1