M5 June 2013 (R) Q2

EdexcelOld spec11 marksMoments Vectors

2. Three forces \(\mathbf{F}_1 = (3\mathbf{i} - \mathbf{j} + \mathbf{k})\) N, \(\mathbf{F}_2 = (2\mathbf{i} - \mathbf{k})\) N, and \(\mathbf{F}_3\) act on a rigid body.

The force \(\mathbf{F}_1\) acts through the point with position vector \((\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) m, the force \(\mathbf{F}_2\) acts through the point with position vector \((\mathbf{i} - 2\mathbf{j})\) m and the force \(\mathbf{F}_3\) acts through the point with position vector \((\mathbf{i} + \mathbf{j} + \mathbf{k})\) m.

Given that the system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) reduces to a couple \(\mathbf{G}\),

(a) find \(\mathbf{G}\). (6)

The line of action of \(\mathbf{F}_3\) is changed so that the system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) now reduces to a couple \((6\mathbf{i} + 8\mathbf{j} + 2\mathbf{k})\) N m.

(b) Find an equation of the new line of action of \(\mathbf{F}_3\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors. (5)

Mark scheme (a)

Scheme	Marks
\(\sum \mathbf{F}_i = \begin{pmatrix}3\\-1\\1\end{pmatrix}+\begin{pmatrix}2\\0\\-1\end{pmatrix}+\mathbf{F}_3 = \begin{pmatrix}0\\0\\0\end{pmatrix}\)	M1
\(\mathbf{F}_3 = \begin{pmatrix}-5\\1\\0\end{pmatrix}\)	A1
\(\mathbf{G} = \begin{pmatrix}1\\2\\1\end{pmatrix}\times\begin{pmatrix}3\\-1\\1\end{pmatrix}+\begin{pmatrix}1\\-2\\0\end{pmatrix}\times\begin{pmatrix}2\\0\\-1\end{pmatrix}+\begin{pmatrix}1\\1\\1\end{pmatrix}\times\begin{pmatrix}-5\\1\\0\end{pmatrix}\)	M1
\(= \begin{pmatrix}3\\2\\-7\end{pmatrix}+\begin{pmatrix}2\\1\\4\end{pmatrix}+\begin{pmatrix}-1\\-5\\6\end{pmatrix}\)	A2
\(= \begin{pmatrix}4\\-2\\3\end{pmatrix}\)	A1
	(6)

Mark scheme (b)

Scheme	Marks
\(\begin{pmatrix}3\\2\\-7\end{pmatrix}+\begin{pmatrix}2\\1\\4\end{pmatrix}+\begin{pmatrix}x\\y\\z\end{pmatrix}\times\begin{pmatrix}-5\\1\\0\end{pmatrix}=\begin{pmatrix}6\\8\\2\end{pmatrix}\)	M1 A1 ft
\(\begin{pmatrix}-z\\-5z\\x + 5y\end{pmatrix}=\begin{pmatrix}1\\5\\5\end{pmatrix}\)	A1
\(z = -1\); put \(x = 0 \Rightarrow y = 1\)	M1
so, \(\mathbf{r} = \begin{pmatrix}0\\1\\-1\end{pmatrix}+t\begin{pmatrix}-5\\1\\0\end{pmatrix}\) is a possible vector equation	A1
	(5)
	(11 marks)