| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding the angle between planes using normal vectors (dot product formula) and finding the line of intersection by solving simultaneous equations. Both parts are textbook exercises with well-established methods, making it slightly easier than average for A-level but typical for Further Maths content. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply a correct normal vector to either plane, e.g. \(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) or \(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\) | B1 | Carry out correct process for evaluating the scalar product of both the normal vectors |
| (ii) Carry out a complete strategy for finding a point on l | M1 | Obtain such a point e.g. \((0, 3, 2)\) |
| EITHER Set up two equations for a direction vector of l, e.g. \(a + 2b - 2c = 0\) and \(2a - 3b + 6c = 0\) | B1 | Solve for one ratio, e.g. \(a:b\) |
| OR Obtain a second point on l, e.g. \((6, -7, -5)\) | A1 | Subtract position vectors to obtain a direction vector for l |
| OR Attempt to find the vector product of the two normal vectors | M1 | Obtain two correct components |
| OR Express one variable in terms of a second | M1 | Obtain a correct simplified expression, e.g. \(x = \frac{9 - 3y}{5}\) |
| OR Express one variable in terms of a second | M1 | Obtain a correct simplified expression, e.g. \(y = \frac{9 - 5x}{3}\) |
**(i)** State or imply a correct normal vector to either plane, e.g. $\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$ or $2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ | B1 | Carry out correct process for evaluating the scalar product of both the normal vectors | M1 | Using the correct process for the moduli, divide the scalar product of the two normals by the product of their moduli and evaluate the inverse cosine of the result | M1 | Obtain answer $40.4°$ or $40.3°$ or $0.705$ (or $0.704$) radians | A1 | [Allow the obtuse answer $139.6°$ or $2.44$ radians] | **[4]**
**(ii)** Carry out a complete strategy for finding a point on l | M1 | Obtain such a point e.g. $(0, 3, 2)$ | A1 |
**EITHER** Set up two equations for a direction vector of l, e.g. $a + 2b - 2c = 0$ and $2a - 3b + 6c = 0$ | B1 | Solve for one ratio, e.g. $a:b$ | M1 | Obtain $a:b:c = 6: -10: -7$, or equivalent | A1 | State a correct answer, e.g. $\mathbf{r} = 3\mathbf{j} + 2\mathbf{k} + \lambda(6\mathbf{i} - 10\mathbf{j} - 7\mathbf{k})$ | A1√ |
**OR** Obtain a second point on l, e.g. $(6, -7, -5)$ | A1 | Subtract position vectors to obtain a direction vector for l | M1 | Obtain $6\mathbf{i} - 10\mathbf{j} - 7\mathbf{k}$, or equivalent | A1 | State a correct answer, e.g. $\mathbf{r} = 3\mathbf{j} + 2\mathbf{k} + \lambda(6\mathbf{i} - 10\mathbf{j} - 7\mathbf{k})$ | A1√ |
**OR** Attempt to find the vector product of the two normal vectors | M1 | Obtain two correct components | A1 | Obtain $6\mathbf{i} - 10\mathbf{j} - 7\mathbf{k}$, or equivalent | A1 | State a correct answer, e.g. $\mathbf{r} = 3\mathbf{j} + 2\mathbf{k} + \lambda(6\mathbf{i} - 10\mathbf{j} - 7\mathbf{k})$ | A1√ |
**OR** Express one variable in terms of a second | M1 | Obtain a correct simplified expression, e.g. $x = \frac{9 - 3y}{5}$ | A1 | Express the same variable in terms of the third and form a three term equation | M1 | Incorporate a correct simplified expression, e.g. $x = \frac{12 - 6z}{7}$ in this equation | A1 | Form a vector equation for the line $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5/3 \\ -7/6 \end{pmatrix}$$ | M1 | State a correct answer, e.g. $\mathbf{r} = 3\mathbf{j} + 2\mathbf{k} + \lambda(-5/3 \mathbf{j} - 7/6 \mathbf{k})$, or equivalent | A1√ |
**OR** Express one variable in terms of a second | M1 | Obtain a correct simplified expression, e.g. $y = \frac{9 - 5x}{3}$ | A1 | Express the third variable in terms of the second | M1 | Obtain a correct simplified expression, e.g. $z = \frac{12 - 7x}{6}$ | A1 | Form a vector equation for the line $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5/3 \\ -7/6 \end{pmatrix}$$ | M1 | State a correct answer, e.g. $\mathbf{r} = 3\mathbf{j} + 2\mathbf{k} + \lambda(-5/3 \mathbf{j} - 7/6 \mathbf{k})$, or equivalent | A1√ | **[6]**9 Two planes have equations $x + 2 y - 2 z = 2$ and $2 x - 3 y + 6 z = 3$. The planes intersect in the straight line $l$.\\
(i) Calculate the acute angle between the two planes.\\
(ii) Find a vector equation for the line $l$.
\hfill \mbox{\textit{CAIE P3 2003 Q9 [10]}}