M5 June 2012 Q5
5. The points \(P\) and \(Q\) have position vectors \(4\mathbf{i} - 6\mathbf{j} - 12\mathbf{k}\) and \(2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\) respectively, relative to a fixed origin \(O\).
Three forces, \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\), act along \(\overrightarrow{OP}\), \(\overrightarrow{QO}\) and \(\overrightarrow{QP}\) respectively, and have magnitudes 7 N, 3 N and \(3\sqrt{10}\) N respectively.
(a) Express \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) in vector form. (3)
(b) Show that the resultant of \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) is \((2\mathbf{i} - 10\mathbf{j} - 16\mathbf{k})\) N. (2)
(c) Find a vector equation of the line of action of this resultant, giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors and \(\lambda\) is a parameter. (5)
| Scheme | Marks |
|---|---|
| \(\mathbf{F}_1 = 7.\dfrac{1}{\sqrt{4^2 + (-6)^2 + (-12)^2}}\begin{pmatrix}4\\-6\\-12\end{pmatrix} = \begin{pmatrix}2\\-3\\-6\end{pmatrix}\) | B1 |
| \(\mathbf{F}_2 = 3.\dfrac{1}{\sqrt{2^2 + 4^2 + 4^2}}\begin{pmatrix}-2\\-4\\-4\end{pmatrix} = \begin{pmatrix}-1\\-2\\-2\end{pmatrix}\) | B1 |
| \(\mathbf{F}_3 = 3\sqrt{10}.\dfrac{1}{\sqrt{2^2 + (-10)^2 + (-16)^2}}\begin{pmatrix}2\\-10\\-16\end{pmatrix} = \begin{pmatrix}1\\-5\\-8\end{pmatrix}\) | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\sum \mathbf{F}_i = \begin{pmatrix}2\\-3\\-6\end{pmatrix} + \begin{pmatrix}-1\\-2\\-2\end{pmatrix} + \begin{pmatrix}1\\-5\\-8\end{pmatrix} = \begin{pmatrix}2\\-10\\-16\end{pmatrix}\) PRINTED ANSWER | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Taking moments about \(O\), \(\begin{pmatrix}4\\-6\\-12\end{pmatrix}\times\begin{pmatrix}1\\-5\\-8\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}\times\begin{pmatrix}2\\-10\\-16\end{pmatrix}\) | M1 |
| \(\begin{pmatrix}-12\\20\\-14\end{pmatrix} = \begin{pmatrix}-16y + 10z\\2z + 16x\\-10x - 2y\end{pmatrix}\) put \(x = 0 \Rightarrow z = 10 \Rightarrow y = 7\) | A1 A1 M1 |
| so, \(\mathbf{r} = \begin{pmatrix}0\\7\\10\end{pmatrix} + \lambda\begin{pmatrix}1\\-5\\-8\end{pmatrix}\) is a vector equation. | A1 |
| (5) | |
| (10 marks) |