| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | December |
| Marks | 10 |
| Topic | Vector Product and Surfaces |
| Type | Vector product calculation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths vector product question with standard calculations and a simple proof. Parts (a)-(b) are routine cross product computations, (c) requires showing λp×F₁=0 using parallel vectors, and (d) follows directly from the definition of zero torque. While it's Further Maths content, the techniques are mechanical with no novel insight required. |
| Spec | 8.04a Vector product: definition, magnitude/direction, component form8.04c Areas using vector product: triangles and parallelograms |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = a \times F_1 = \begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 6 \\ 7 \\ -3 \end{pmatrix} = \begin{pmatrix} -19 \\ 15 \\ -3 \end{pmatrix}\) | M1, A1 | 3.4, 1.1 |
| Total: [2] | Attempt at cross-product, soi; BC |
| Answer | Marks | Guidance |
|---|---|---|
| \(b \times F_2 = \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix} \times \begin{pmatrix} -7 \\ -10 \\ 2 \end{pmatrix} = \begin{pmatrix} 20 \\ -13 \\ 5 \end{pmatrix}\) | M1 | 3.4 |
| Calculating magnitudes: | ||
| \(\ | b \times F_2\ | = \sqrt{594}\) |
| \(\ | a \times F_1\ | = \sqrt{595}\) so \(F_1\) exerts the torque of greater magnitude |
| Total: [3] | soi; BC; BC |
| Answer | Marks | Guidance |
|---|---|---|
| \((a + \lambda p) \times F_1 = a \times F_1 + \lambda p \times F_1\) | M1 | 3.1a |
| but \(p\) and \(F\) are parallel implies \(p \times F_1 = 0\) | B1 | 2.2a |
| So \((a + \lambda p) \times F_1 = a \times F_1\) as in part (a) | A1 | 1.1 |
| Total: [3] | Use distributive property; use parallel vectors property in the correct direction; conclude |
| Answer | Marks | Guidance |
|---|---|---|
| \(F_1\) exerts no torque means that \(a \times F_1 = 0\) | M1 | 3.4 |
| Vector product = 0 implies that \(a\) and \(F_1\) are parallel, so \(F_1\) acts in the same direction as the line \(OA\) | E1 | 2.2a |
| Total: [2] | Must not just verify; use parallel vectors property in the correct direction |
**Part (a)**
| $T = a \times F_1 = \begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 6 \\ 7 \\ -3 \end{pmatrix} = \begin{pmatrix} -19 \\ 15 \\ -3 \end{pmatrix}$ | M1, A1 | 3.4, 1.1 |
| Total: [2] | Attempt at cross-product, soi; BC | |
**Part (b)**
| $b \times F_2 = \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix} \times \begin{pmatrix} -7 \\ -10 \\ 2 \end{pmatrix} = \begin{pmatrix} 20 \\ -13 \\ 5 \end{pmatrix}$ | M1 | 3.4 |
| Calculating magnitudes: | | |
| $\|b \times F_2\| = \sqrt{594}$ | A1 | 1.1 |
| $\|a \times F_1\| = \sqrt{595}$ so $F_1$ exerts the torque of greater magnitude | A1 | 1.1 |
| Total: [3] | soi; BC; BC | |
**Part (c)**
| $(a + \lambda p) \times F_1 = a \times F_1 + \lambda p \times F_1$ | M1 | 3.1a |
| but $p$ and $F$ are parallel implies $p \times F_1 = 0$ | B1 | 2.2a |
| So $(a + \lambda p) \times F_1 = a \times F_1$ as in part (a) | A1 | 1.1 |
| Total: [3] | Use distributive property; use parallel vectors property in the correct direction; conclude | |
**Part (d)**
| $F_1$ exerts no torque means that $a \times F_1 = 0$ | M1 | 3.4 |
| Vector product = 0 implies that $a$ and $F_1$ are parallel, so $F_1$ acts in the same direction as the line $OA$ | E1 | 2.2a |
| Total: [2] | Must not just verify; use parallel vectors property in the correct direction | |5 Torque is a vector quantity that measures how much a force acting on an object causes that object to rotate.
The torque (about the origin), $\mathbf { T }$, exerted on an object is given by $\mathbf { T } = \mathbf { p } \times \mathbf { F }$, where $\mathbf { F }$ is the force acting on the object and $\mathbf { p }$ is the position vector of the point at which $\mathbf { F }$ is applied to the object.
The points $A$ and $B$, with position vectors $\mathbf { a } = 3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }$ and $\mathbf { b } = 3 \mathbf { i } + 5 \mathbf { j } + \mathbf { k }$ are on the surface of a rock. The force $\mathbf { F } _ { 1 } = 6 \mathbf { i } + 7 \mathbf { j } - 3 \mathbf { k }$ is applied to the rock at $A$ while the force $\mathbf { F } _ { 2 } = - 7 \mathbf { i } - 10 \mathbf { j } + 2 \mathbf { k }$ is applied to the rock at $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the torque (about the origin) exerted on the rock by $\mathbf { F } _ { 1 }$.
\item Determine which of the two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ exerts a torque (about the origin) of greater magnitude on the rock.
\item Show that the torque (about the origin) is the same as your answer to part (a) when $\mathbf { F } _ { 1 }$ acts on the rock at any point on the line $\mathbf { r } = \mathbf { a } + \lambda \mathbf { p }$, where $\mathbf { p }$ is a vector in the same direction as $\mathbf { F } _ { 1 }$.
A third force $\mathbf { F } _ { 3 }$ is now applied to the rock at $A$, which exerts zero torque (about the origin).
\item Show that $\mathbf { F } _ { 3 }$ must act in the direction of the line through $A$ and the origin.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q5 [10]}}