M1 June 2012 Q6

EdexcelOld spec13 marksVectors

6. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.]

A ship \(S\) is moving with constant velocity \((-12\mathbf{i} + 7.5\mathbf{j})\) km h\(^{-1}\).

(a) Find the direction in which \(S\) is moving, giving your answer as a bearing. (3)

At time \(t\) hours after noon, the position vector of \(S\) is \(\mathbf{s}\) km. When \(t = 0\), \(\mathbf{s} = 40\mathbf{i} - 6\mathbf{j}\).

(b) Write down \(\mathbf{s}\) in terms of \(t\). (2)

A fixed beacon \(B\) is at the point with position vector \((7\mathbf{i} + 12.5\mathbf{j})\) km.

(d) Find the distance of \(S\) from \(B\) when \(S\) is due north of \(B\). (4)

Mark scheme (a)

Scheme	Marks
\(\arctan\dfrac{7.5}{12} = 32^\circ\)	M1 A1
Bearing is 302 (allow more accuracy)	A1
	(3)

Notes

First M1 for \(\arctan\left(\dfrac{\pm 7.5}{\pm 12}\right)\) either way up

First A1 for a correct value from their expression, usually 32\(^\circ\) or 58\(^\circ\)

Second A1 for 302 (allow more accurate answers)

Mark scheme (b)

Scheme	Marks
\(\mathbf{s} = 40\mathbf{i} - 6\mathbf{j} + t(-12\mathbf{i} + 7.5\mathbf{j})\)	M1 A1
	(2)

Notes

M1 for a clear attempt at \((40\mathbf{i} - 6\mathbf{j}) + t(-12\mathbf{i} + 7.5\mathbf{j})\)

A1 for any correct expression

Mark scheme (c)

Scheme	Marks
\(t = 3\), \(\mathbf{s} = 4\mathbf{i} + 16.5\mathbf{j}\)	M1
\(\mathbf{s} - \mathbf{b} = -3\mathbf{i} + 4\mathbf{j}\)	M1
\(SB = \sqrt{\left((-3)^2 + 4^2\right)} = 5\) (km)	DM1 A1
	(4)

Notes

First M1 is really B1 for \(4\mathbf{i} + 16.5\mathbf{j}\) (seen or implied but can be in unsimplified form)

Second M1 is for a subtraction, \(\mathbf{s} - \mathbf{b}\) or \(\mathbf{b} - \mathbf{s}\).

Third DM1, dependent on second M1, for finding magnitude of their \(\mathbf{s} - \mathbf{b}\) or \(\mathbf{b} - \mathbf{s}\)

A1 for 5

Mark scheme (d)

Scheme	Marks
Equating \(\mathbf{i}\) components \(40 - 12t = 7\) or \(-33 + 12t = 0\)	M1
\(t = 2\dfrac{3}{4}\)	A1
When \(t = 2\dfrac{3}{4}\), \(\mathbf{s} = (7\mathbf{i}) + 14\dfrac{5}{8}\mathbf{j}\)	M1
\(SB = 2\dfrac{1}{8}\) (km) 2.125, 2.13	A1
	(4)
	(13 marks)

Notes

First M1 for equating \(\mathbf{i}\)-component of their answer in part (b) to 7 or the \(\mathbf{i}\)-component of their \(\mathbf{s} - \mathbf{b}\) or \(\mathbf{b} - \mathbf{s}\) to zero

First A1 for 2.75 cao

Second M1 (independent) for attempt to find \(\mathbf{j}\)-component of their \(\mathbf{s}\) at their \(t = 2.75\)

Second A1 2.125 or 2.13 cao

OR

When \(t = 2\dfrac{3}{4}\), \(7.5t - 18.5 = 2.125,\ 2.13\)

M1 A1