M4 June 2014 Q1

EdexcelOld spec8 marksVectors

1. A particle \(A\) has constant velocity \((3\mathbf{i} + \mathbf{j})\) m s\(^{-1}\) and a particle \(B\) has constant velocity \((\mathbf{i} - \mathbf{k})\) m s\(^{-1}\). At time \(t = 0\) seconds, the position vectors of the particles \(A\) and \(B\) with respect to a fixed origin \(O\) are \((-6\mathbf{i} + 4\mathbf{j} - 3\mathbf{k})\) m and \((-2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\) m respectively.

(a) Show that, in the subsequent motion, the minimum distance between \(A\) and \(B\) is \(4\sqrt{2}\) m. (6)

(b) Find the position vector of \(A\) at the instant when the distance between \(A\) and \(B\) is a minimum. (2)

Mark scheme (a)

Scheme	Marks
\(\mathbf{r}_A = (-6\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}) + t(3\mathbf{i} + \mathbf{j}) = ((-6 + 3t)\mathbf{i} + (4 + t)\mathbf{j} + (-3)\mathbf{k})\)	M1
\(\mathbf{r}_B = (-2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + t(\mathbf{i} - \mathbf{k}) = ((-2 + t)\mathbf{i} + (2)\mathbf{j} + (3 - t)\mathbf{k})\)	A1
\({}_B\mathbf{r}_A = (-2 + t + 6 - 3t)\mathbf{i} + (2 - 4 - t)\mathbf{j} + (3 - t + 3)\mathbf{k}\) \(\quad = (4 - 2t)\mathbf{i} + (-2 - t)\mathbf{j} + (6 - t)\mathbf{k}\)	M1
\(\left\|{}_B\mathbf{r}_A\right\|^2 = (4 - 2t)^2 + (t + 2)^2 + (6 - t)^2\)	M1
alt1 \(\quad = 6t^2 - 24t + 56 = 6(t - 2)^2 + 32\)	M1
Minimum distance \(= \sqrt{32} = 4\sqrt{2}\) m **	A1
	(6)

Notes

M1 Position vector for \(A\) or \(B\)

A1 Both position vectors correct (seen or implied)

M1 Position of \(B\) relative to \(A\) (or \(A\) relative to \(B\))

M1 Use of Pythagoras

M1 Complete the square

A1 Reach given answer correctly

alt2

\(\left\|{}_B\mathbf{r}_A\right\|^2 = (4 - 2t)^2 + (t + 2)^2 + (6 - t)^2\ \left(= 6t^2 - 24t + 56\right)\)	M1
\(12t - 24 = 0 \Rightarrow t = 2\)	M1
Minimum distance \(= \sqrt{32} = 4\sqrt{2}\) m **	A1

M1 Use of Pythagoras

M1 Differentiate and solve for \(t\)

A1 Reach given answer correctly

alt3

\(\begin{pmatrix}4 - 2t\\-2 - t\\6 - t\end{pmatrix}\bullet\begin{pmatrix}2\\1\\1\end{pmatrix} = 0 \Rightarrow 8 - 4t - 2 - t + 6 - t = 12 - 6t = 0\)	M1
Distance \(= \sqrt{0^2 + 4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\)	M1
	A1

M1 Scalar product of position vector with relative velocity = zero and form equation in \(t\)

M1 Use of Pythagoras

A1 Reach given answer correctly

Scheme	Marks
When \(t = 2\),	B1
\(\mathbf{r}_A = 6\mathbf{j} - 3\mathbf{k}\)	B1
	(2)
	(8 marks)