| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding a midpoint, recognizing that PQ is perpendicular to the plane (giving the normal vector), writing the Cartesian equation, then finding an intersection point and calculating distance. All steps are routine applications of A-level formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply \(A\) is \((1, 4, -2)\) | B1 | |
| State or imply \(\overrightarrow{QP} = 12\mathbf{i} + 6\mathbf{j} - 6\mathbf{k}\) or equivalent | B1 | |
| Use \(QP\) as normal and \(A\) as mid-point to find equation of plane | M1 | |
| Obtain \(12x + 6y - 6z = 48\) or equivalent | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: State equation of \(PB\) is \(\mathbf{r} = 7\mathbf{i} + 7\mathbf{j} - 5\mathbf{k} + \lambda\mathbf{i}\) | B1 | |
| Set up and solve a relevant equation for \(\lambda\) | M1 | |
| Obtain \(\lambda = -9\) and hence \(B\) is \((-2, 7, -5)\) | A1 | |
| Use correct method to find distance between \(A\) and \(B\) | M1 | |
| Obtain \(5.20\) | A1 | |
| Or: Obtain \(12\) for result of scalar product of \(QP\) and \(\mathbf{i}\) or equivalent | B1 | |
| Use correct method involving moduli, scalar product and cosine to find angle \(APB\) | M1 | |
| Obtain \(35.26°\) or equivalent | A1 | |
| Use relevant trigonometry to find \(AB\) | M1 | |
| Obtain \(5.20\) | A1 | [5] |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $A$ is $(1, 4, -2)$ | B1 | |
| State or imply $\overrightarrow{QP} = 12\mathbf{i} + 6\mathbf{j} - 6\mathbf{k}$ or equivalent | B1 | |
| Use $QP$ as normal and $A$ as mid-point to find equation of plane | M1 | |
| Obtain $12x + 6y - 6z = 48$ or equivalent | A1 | [4] |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** State equation of $PB$ is $\mathbf{r} = 7\mathbf{i} + 7\mathbf{j} - 5\mathbf{k} + \lambda\mathbf{i}$ | B1 | |
| Set up and solve a relevant equation for $\lambda$ | M1 | |
| Obtain $\lambda = -9$ and hence $B$ is $(-2, 7, -5)$ | A1 | |
| Use correct method to find distance between $A$ and $B$ | M1 | |
| Obtain $5.20$ | A1 | |
| **Or:** Obtain $12$ for result of scalar product of $QP$ and $\mathbf{i}$ or equivalent | B1 | |
| Use correct method involving moduli, scalar product and cosine to find angle $APB$ | M1 | |
| Obtain $35.26°$ or equivalent | A1 | |
| Use relevant trigonometry to find $AB$ | M1 | |
| Obtain $5.20$ | A1 | [5] |
---6 The points $P$ and $Q$ have position vectors, relative to the origin $O$, given by
$$\overrightarrow { O P } = 7 \mathbf { i } + 7 \mathbf { j } - 5 \mathbf { k } \quad \text { and } \quad \overrightarrow { O Q } = - 5 \mathbf { i } + \mathbf { j } + \mathbf { k }$$
The mid-point of $P Q$ is the point $A$. The plane $\Pi$ is perpendicular to the line $P Q$ and passes through $A$.\\
(i) Find the equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.\\
(ii) The straight line through $P$ parallel to the $x$-axis meets $\Pi$ at the point $B$. Find the distance $A B$, correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P3 2013 Q6 [9]}}