| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard three-part vectors question covering routine techniques: (a) finding a line equation from two points (direct subtraction), (b) angle between lines using dot product formula, and (c) showing non-intersection by solving simultaneous equations. All are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain direction vector \(-\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) | B1 | OE |
| Use a correct method to form a vector equation | M1 | |
| Obtain answer \(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) or \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) | A1 | Need r or r on LHS |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Carry out correct process for evaluating the scalar product of the direction vectors | M1 | \((-1, -3, 1) \cdot (1, -3, -2) = -1 + 9 - 2\) |
| Using correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors | M1 | \(\cos^{-1}\left(\frac{1 + 9 - 2}{(1+9+1)(1+9+4)}\right)\) |
| Obtain answer \(61.1°\) | A1 | \(61.086°\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express general point of \(AB\) or \(l\) in component form, e.g. \((2-\lambda, 1-3\lambda, 1+\lambda)\) or \((1+\mu, 2-3\mu, -3-2\mu)\) | B1 | |
| Equate at least two pairs of components and solve for \(\lambda\) or for \(\mu\) | M1 | |
| Obtain a correct answer for \(\lambda\) or \(\mu\), e.g. \(\lambda = 6, \frac{1}{3}\), or \(-\frac{14}{9}\); \(\mu = -5, \frac{2}{3}\) or \(-\frac{11}{9}\) | A1 | |
| Verify that all three equations are not satisfied, and the lines do not intersect | A1 | |
| Express general point of \(AB\) or \(l\) in component form, e.g. \((1-\lambda^*, -2-3\lambda^*, 2+\lambda^*)\) or \((1+\mu^*, 2-3\mu^*, -3-2\mu^*)\) | 4 |
## Question 10:
### Part 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain direction vector $-\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ | B1 | OE |
| Use a correct method to form a vector equation | M1 | |
| Obtain answer $\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})$ or $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})$ | A1 | Need **r** or r on LHS |
| | **3** | |
### Part 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Carry out correct process for evaluating the scalar product of the direction vectors | M1 | $(-1, -3, 1) \cdot (1, -3, -2) = -1 + 9 - 2$ |
| Using correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors | M1 | $\cos^{-1}\left(\frac{1 + 9 - 2}{(1+9+1)(1+9+4)}\right)$ |
| Obtain answer $61.1°$ | A1 | $61.086°$ |
| | **3** | |
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Express general point of $AB$ or $l$ in component form, e.g. $(2-\lambda, 1-3\lambda, 1+\lambda)$ or $(1+\mu, 2-3\mu, -3-2\mu)$ | **B1** | |
| Equate at least two pairs of components and solve for $\lambda$ or for $\mu$ | **M1** | |
| Obtain a correct answer for $\lambda$ or $\mu$, e.g. $\lambda = 6, \frac{1}{3}$, or $-\frac{14}{9}$; $\mu = -5, \frac{2}{3}$ or $-\frac{11}{9}$ | **A1** | |
| Verify that all three equations are not satisfied, and the lines do not intersect | **A1** | |
| Express general point of $AB$ or $l$ in component form, e.g. $(1-\lambda^*, -2-3\lambda^*, 2+\lambda^*)$ or $(1+\mu^*, 2-3\mu^*, -3-2\mu^*)$ | **4** | |
---10 The points $A$ and $B$ have position vectors $2 \mathbf { i } + \mathbf { j } + \mathbf { k }$ and $\mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k }$ respectively. The line $l$ has vector equation $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k } + \mu ( \mathbf { i } - 3 \mathbf { j } - 2 \mathbf { k } )$.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for the line through $A$ and $B$.
\item Find the acute angle between the directions of $A B$ and $l$, giving your answer in degrees.
\item Show that the line through $A$ and $B$ does not intersect the line $l$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q10 [10]}}