| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Moderate -0.3 This is a straightforward application of standard vector methods: finding a plane equation using a normal vector (direction of AB) and a point, then calculating an angle using dot product. Both parts require routine techniques with no conceptual challenges, making it slightly easier than average but still requiring proper execution of multiple steps. |
| 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 forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain \(\pm\begin{pmatrix}3\\-4\\6\end{pmatrix}\) as normal to plane | B1 | |
| Form equation of \(p\) as \(3x - 4y + 6z = k\) or \(-3x + 4y - 6z = k\) and use relevant point to find \(k\) | M1 | |
| Obtain \(3x - 4y + 6z = 80\) or \(-3x + 4y - 6z = -80\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State direction vector \(\begin{pmatrix}0\\1\\0\end{pmatrix}\) or equivalent | B1 | |
| Carry out correct process for finding scalar product of two relevant vectors | M1 | |
| Use correct complete process with moduli and scalar product and evaluate \(\sin^{-1}\) or \(\cos^{-1}\) of result | M1 | |
| Obtain \(30.8°\) or \(0.538\) radians | A1 | [4] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain $\pm\begin{pmatrix}3\\-4\\6\end{pmatrix}$ as normal to plane | B1 | |
| Form equation of $p$ as $3x - 4y + 6z = k$ or $-3x + 4y - 6z = k$ and use relevant point to find $k$ | M1 | |
| Obtain $3x - 4y + 6z = 80$ or $-3x + 4y - 6z = -80$ | A1 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State direction vector $\begin{pmatrix}0\\1\\0\end{pmatrix}$ or equivalent | B1 | |
| Carry out correct process for finding scalar product of two relevant vectors | M1 | |
| Use correct complete process with moduli and scalar product and evaluate $\sin^{-1}$ or $\cos^{-1}$ of result | M1 | |
| Obtain $30.8°$ or $0.538$ radians | A1 | [4] |
---3 Points $A$ and $B$ have coordinates $( - 1,2,5 )$ and $( 2 , - 2,11 )$ respectively. The plane $p$ passes through $B$ and is perpendicular to $A B$.\\
(i) Find an equation of $p$, giving your answer in the form $a x + b y + c z = d$.\\
(ii) Find the acute angle between $p$ and the $y$-axis.
\hfill \mbox{\textit{CAIE P3 2011 Q3 [7]}}