| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.8 This is a two-part vectors question requiring (i) solving a quadratic equation from the distance formula applied to a parametric line, and (ii) using the relationship between line-plane angles and dot products to find unknown plane coefficients. Part (ii) requires understanding that the angle between a line and plane relates to the angle between the direction vector and normal vector, then solving an equation involving the given sine value. This goes beyond routine application and requires synthesis of multiple vector concepts, placing it moderately above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Express general point of \(l\) in component form, e.g. \((1+2\lambda,\ 2-\lambda,\ 1+\lambda)\) | B1 | |
| Using correct process for the modulus form an equation in \(\lambda\) | M1* | |
| Reduce the equation to a quadratic, e.g. \(6\lambda^2 + 2\lambda - 4 = 0\) | A1 | |
| Solve for \(\lambda\) (usual requirements for solution of quadratic) | DM1 | |
| Obtain final answers \(-\mathbf{i}+3\mathbf{j}\) and \(\frac{7}{3}\mathbf{i}+\frac{4}{3}\mathbf{j}+\frac{5}{3}\mathbf{k}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using correct process, find scalar product of direction vector of \(l\) and normal for \(p\) | M1 | |
| Using correct process for moduli, divide scalar product by product of moduli and equate to \(\frac{2}{3}\) | M1 | |
| State correct equation in any form, e.g. \(\frac{2a-1+1}{\sqrt{(a^2+1+1)}\cdot\sqrt{(2^2+(-1)^2+1)}} = \pm\frac{2}{3}\) | A1 | |
| Solve for \(a^2\) | M1 | |
| Obtain answer \(a = \pm 2\) | A1 | [5] |
## Question 10:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Express general point of $l$ in component form, e.g. $(1+2\lambda,\ 2-\lambda,\ 1+\lambda)$ | B1 | |
| Using correct process for the modulus form an equation in $\lambda$ | M1* | |
| Reduce the equation to a quadratic, e.g. $6\lambda^2 + 2\lambda - 4 = 0$ | A1 | |
| Solve for $\lambda$ (usual requirements for solution of quadratic) | DM1 | |
| Obtain final answers $-\mathbf{i}+3\mathbf{j}$ and $\frac{7}{3}\mathbf{i}+\frac{4}{3}\mathbf{j}+\frac{5}{3}\mathbf{k}$ | A1 | [5] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using correct process, find scalar product of direction vector of $l$ and normal for $p$ | M1 | |
| Using correct process for moduli, divide scalar product by product of moduli and equate to $\frac{2}{3}$ | M1 | |
| State correct equation in any form, e.g. $\frac{2a-1+1}{\sqrt{(a^2+1+1)}\cdot\sqrt{(2^2+(-1)^2+1)}} = \pm\frac{2}{3}$ | A1 | |
| Solve for $a^2$ | M1 | |
| Obtain answer $a = \pm 2$ | A1 | [5] |10 The line $l$ has vector equation $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \lambda ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$.\\
(i) Find the position vectors of the two points on the line whose distance from the origin is $\sqrt { } ( 10 )$.\\
(ii) The plane $p$ has equation $a x + y + z = 5$, where $a$ is a constant. The acute angle between the line $l$ and the plane $p$ is equal to $\sin ^ { - 1 } \left( \frac { 2 } { 3 } \right)$. Find the possible values of $a$.
\hfill \mbox{\textit{CAIE P3 2016 Q10 [10]}}