| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard two-part vectors question requiring routine techniques: part (a) involves equating components and solving simultaneous equations to find the intersection condition (standard procedure), while part (b) uses the scalar product formula for angles between lines. Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express general point of at least one line correctly in component form, i.e. \((1+a\lambda,\ 2+2\lambda,\ 1-\lambda)\) or \((2+2\mu,\ 1-\mu,\ -1+\mu)\) | B1 | |
| Equate at least two pairs of corresponding components and solve for \(\lambda\) or for \(\mu\) | M1 | May be implied: \(1+a\lambda=2+2\mu\), \(2+2\lambda=1-\mu\), \(1-\lambda=-1+\mu\) |
| Obtain \(\lambda = -3\) or \(\mu = 5\) | A1 | |
| Obtain \(a = -\frac{11}{3}\) | A1 | Allow \(a = -3.667\) |
| State that the point of intersection has position vector \(12\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\) | A1 | Allow coordinate form \((12, -4, 4)\) |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct process for finding the scalar product of direction vectors for the two lines | M1 | \((a, 2, -1)\cdot(2,-1,1) = 2a-2-1\) or \(2a-3\) |
| Using correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to \(\pm\frac{1}{6}\) | \*M1 | |
| State a correct equation in \(a\) in any form, e.g. \(\frac{2a-2-1}{\sqrt{6}\sqrt{(a^2+5)}} = \pm\frac{1}{6}\) | A1 | |
| Solve for \(a\) | DM1 | Solve 3-term quadratic for \(a\) having expanded \((2a-3)^2\); \(36(2a-3)^2 = 6(a^2+5)\); \(138a^2 - 432a + 294 = 0\); \(23a^2 - 72a + 49 = 0\); \((23a-49)(a-1)=0\) |
| Obtain \(a = 1\) | A1 | |
| Obtain \(a = \frac{49}{23}\) | A1 | Allow \(a = 2.13\) |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos(\theta) = \frac{[ | a^2+2^2+(-1)^2 | ^2 + |
| Equate the result to \(\pm\frac{1}{6}\) | \*M1 A1 | Allow M1\* here for any two vectors |
| Solve for \(a\) | DM1 | As above |
| Obtain \(a = 1\) | A1 | |
| Obtain \(a = \frac{49}{23}\) | A1 | Allow \(a = 2.13\) |
| Total | 6 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Express general point of at least one line correctly in component form, i.e. $(1+a\lambda,\ 2+2\lambda,\ 1-\lambda)$ or $(2+2\mu,\ 1-\mu,\ -1+\mu)$ | B1 | |
| Equate at least two pairs of corresponding components and solve for $\lambda$ or for $\mu$ | M1 | May be implied: $1+a\lambda=2+2\mu$, $2+2\lambda=1-\mu$, $1-\lambda=-1+\mu$ |
| Obtain $\lambda = -3$ or $\mu = 5$ | A1 | |
| Obtain $a = -\frac{11}{3}$ | A1 | Allow $a = -3.667$ |
| State that the point of intersection has position vector $12\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$ | A1 | Allow coordinate form $(12, -4, 4)$ |
| **Total** | **5** | |
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## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct process for finding the scalar product of direction vectors for the two lines | M1 | $(a, 2, -1)\cdot(2,-1,1) = 2a-2-1$ or $2a-3$ |
| Using correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to $\pm\frac{1}{6}$ | \*M1 | |
| State a correct equation in $a$ in any form, e.g. $\frac{2a-2-1}{\sqrt{6}\sqrt{(a^2+5)}} = \pm\frac{1}{6}$ | A1 | |
| Solve for $a$ | DM1 | Solve 3-term quadratic for $a$ having expanded $(2a-3)^2$; $36(2a-3)^2 = 6(a^2+5)$; $138a^2 - 432a + 294 = 0$; $23a^2 - 72a + 49 = 0$; $(23a-49)(a-1)=0$ |
| Obtain $a = 1$ | A1 | |
| Obtain $a = \frac{49}{23}$ | A1 | Allow $a = 2.13$ |
| **Total** | **6** | |
**Alternative method for 11(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos(\theta) = \frac{[|a^2+2^2+(-1)^2|^2 + |2^2+(-1)^2+1^2|}{-|(a-2)^2+3^2+(-2)^2|^2]/[2|a^2+2^2+(-1)^2|\cdot|2^2+(-1)^2+1^2|]}$ | M1 | Use of cosine rule. Must be correct vectors. |
| Equate the result to $\pm\frac{1}{6}$ | \*M1 A1 | Allow M1\* here for any two vectors |
| Solve for $a$ | DM1 | As above |
| Obtain $a = 1$ | A1 | |
| Obtain $a = \frac{49}{23}$ | A1 | Allow $a = 2.13$ |
| **Total** | **6** | |11 Two lines have equations $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \lambda ( a \mathbf { i } + 2 \mathbf { j } - \mathbf { k } )$ and $\mathbf { r } = 2 \mathbf { i } + \mathbf { j } - \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$, where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Given that the two lines intersect, find the value of $a$ and the position vector of the point of intersection.
\item Given instead that the acute angle between the directions of the two lines is $\cos ^ { - 1 } \left( \frac { 1 } { 6 } \right)$, find the two possible values of $a$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q11 [11]}}