| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| 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 two-part vectors question requiring routine techniques: finding the angle between planes using the dot product of normal vectors, then finding the line of intersection by solving simultaneous equations and using the cross product for direction. While it involves multiple steps, both parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply a correct normal vector to either plane, e.g. \(\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\), or \(2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) | B1 | |
| Carry out correct process for evaluating the scalar product of two 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 \(85.9°\) or 1.50 radians | A1 | 4 |
| (ii) EITHER: Carry out a complete strategy for finding a point on \(l\) | M1 | |
| Obtain such a point, e.g. (0, 2, 1) | A1 | |
| EITHER: State two equations for a direction vector \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) for \(l\), e.g. \(a + 3b - 2c = 0\) and \(2a + b + 3c = 0\) | B1 | |
| Solve for one ratio, e.g. \(a : b\) | M1 | |
| Obtain \(a : b : c = 11 : -7 : -5\) | A1 | |
| State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k})\) | A1* | |
| OR1: Obtain a second point on \(l\), e.g. \(\left(\frac{22}{7}, 0, -\frac{3}{7}\right)\) | B1 | |
| Subtract position vectors and obtain a direction vector for \(l\) | M1 | |
| Obtain \(22\mathbf{i} - 14\mathbf{j} - 10\mathbf{k}\), or equivalent | A1 | |
| State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(22\mathbf{i} - 14\mathbf{j} - 10\mathbf{k})\) | A1* | |
| OR2: Attempt to find the vector product of the two normal vectors | M1 | |
| Obtain two correct components | A1 | |
| Obtain \(11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k}\), or equivalent | A1 | |
| State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k})\) | A1* | |
| OR3: Express one variable in terms of a second | M1 | |
| Obtain a correct simplified expression, e.g. \(x = (22 - 11y)/7\) | A1 | |
| Express the same variable in terms of the third | M1 | |
| Obtain a correct simplified expression, e.g. \(x = (11 - 11z)/5\) | A1 | |
| Form a vector equation for the line | M1 | |
| State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda\left(\mathbf{i} - \frac{7}{11}\mathbf{j} - \frac{5}{11}\mathbf{k}\right)\) | A1* | 6 |
**(i)** State or imply a correct normal vector to either plane, e.g. $\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$, or $2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ | B1 |
Carry out correct process for evaluating the scalar product of two 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 $85.9°$ or 1.50 radians | A1 | 4 |
**(ii)** **EITHER:** Carry out a complete strategy for finding a point on $l$ | M1 |
Obtain such a point, e.g. (0, 2, 1) | A1 |
**EITHER:** State two equations for a direction vector $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ for $l$, e.g. $a + 3b - 2c = 0$ and $2a + b + 3c = 0$ | B1 |
Solve for one ratio, e.g. $a : b$ | M1 |
Obtain $a : b : c = 11 : -7 : -5$ | A1 |
State a correct answer, e.g. $\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k})$ | A1* |
**OR1:** Obtain a second point on $l$, e.g. $\left(\frac{22}{7}, 0, -\frac{3}{7}\right)$ | B1 |
Subtract position vectors and obtain a direction vector for $l$ | M1 |
Obtain $22\mathbf{i} - 14\mathbf{j} - 10\mathbf{k}$, or equivalent | A1 |
State a correct answer, e.g. $\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(22\mathbf{i} - 14\mathbf{j} - 10\mathbf{k})$ | A1* |
**OR2:** Attempt to find the vector product of the two normal vectors | M1 |
Obtain two correct components | A1 |
Obtain $11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k}$, or equivalent | A1 |
State a correct answer, e.g. $\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda(11\mathbf{i} - 7\mathbf{j} - 5\mathbf{k})$ | A1* |
**OR3:** Express one variable in terms of a second | M1 |
Obtain a correct simplified expression, e.g. $x = (22 - 11y)/7$ | A1 |
Express the same variable in terms of the third | M1 |
Obtain a correct simplified expression, e.g. $x = (11 - 11z)/5$ | A1 |
Form a vector equation for the line | M1 |
State a correct answer, e.g. $\mathbf{r} = 2\mathbf{j} + \mathbf{k} + \lambda\left(\mathbf{i} - \frac{7}{11}\mathbf{j} - \frac{5}{11}\mathbf{k}\right)$ | A1* | 6 |
[The ✓ marks are dependent on all M marks being earned.]9 Two planes have equations $x + 3 y - 2 z = 4$ and $2 x + y + 3 z = 5$. 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 2015 Q9 [10]}}