| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: writing a line equation, finding a parameter by substituting a point, and using the scalar product formula for angles between lines. Part (c) requires solving a quadratic but involves no novel insight. Slightly easier than average A-level due to routine application of formulas. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use a correct method to form a vector equation | M1 | Allow in column vectors. |
| Obtain \(\mathbf{r}=8\mathbf{i}-5\mathbf{j}+6\mathbf{k}+\lambda(2\mathbf{i}+\mathbf{j}+4\mathbf{k})\) | A1 | Need \(\mathbf{r}=\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State the position vector of a point on \(l\) in component form. Or at least 2 correct components seen | B1 FT | Follow *their* equation \((8+2\lambda)\mathbf{i}+(-5+\lambda)\mathbf{j}+(6+4\lambda)\mathbf{k}\). Might see the correct equation for the first time in (b). |
| Equate to \(-n\mathbf{i}+4t\mathbf{j}+3t\mathbf{k}\) and solve for \(t\) | M1 | |
| Obtain \(t=-2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Evaluate the scalar product of a pair of relevant vectors | M1 | \((2\mathbf{i}+\mathbf{j}+4\mathbf{k})\cdot(a\mathbf{i}-\mathbf{j}+3\mathbf{k})=2a+11\) OE, SOI |
| Complete the process for finding the cosine of \(\theta\) | *M1 | Divide the scalar product by the product of the moduli and equate to \(\cos\theta\). |
| Obtain \(\frac{2a+11}{\sqrt{21}\sqrt{10+a^2}}=\pm\frac{1}{\sqrt{6}}\) | A1 | OE |
| Form a 3-term quadratic equation in \(a\) and solve for \(a\) | DM1 | \(a^2+88a+172=0\) OE |
| Obtain \(a=-2\), \(a=-86\) | A1 | Correct only (both values). |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use a correct method to form a vector equation | M1 | Allow in column vectors. |
| Obtain $\mathbf{r}=8\mathbf{i}-5\mathbf{j}+6\mathbf{k}+\lambda(2\mathbf{i}+\mathbf{j}+4\mathbf{k})$ | A1 | Need $\mathbf{r}=\ldots$ |
**Total: 2 marks**
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State the position vector of a point on $l$ in component form. Or at least 2 correct components seen | B1 FT | Follow *their* equation $(8+2\lambda)\mathbf{i}+(-5+\lambda)\mathbf{j}+(6+4\lambda)\mathbf{k}$. Might see the correct equation for the first time in **(b)**. |
| Equate to $-n\mathbf{i}+4t\mathbf{j}+3t\mathbf{k}$ and solve for $t$ | M1 | |
| Obtain $t=-2$ | A1 | |
**Total: 3 marks**
## Question 9(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Evaluate the scalar product of a pair of relevant vectors | M1 | $(2\mathbf{i}+\mathbf{j}+4\mathbf{k})\cdot(a\mathbf{i}-\mathbf{j}+3\mathbf{k})=2a+11$ OE, SOI |
| Complete the process for finding the cosine of $\theta$ | *M1 | Divide the scalar product by the product of the moduli and equate to $\cos\theta$. |
| Obtain $\frac{2a+11}{\sqrt{21}\sqrt{10+a^2}}=\pm\frac{1}{\sqrt{6}}$ | A1 | OE |
| Form a 3-term quadratic equation in $a$ and solve for $a$ | DM1 | $a^2+88a+172=0$ OE |
| Obtain $a=-2$, $a=-86$ | A1 | Correct only (both values). |
**Total: 5 marks**
---9 The position vector of point $A$ relative to the origin $O$ is $\overrightarrow { O A } = 8 \mathbf { i } - 5 \mathbf { j } + 6 \mathbf { k }$.\\
The line $l$ passes through $A$ and is parallel to the vector $2 \mathbf { i } + \mathbf { j } + 4 \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item State a vector equation for $l$.
\item The position vector of point $B$ relative to the origin $O$ is $\overrightarrow { O B } = - t \mathbf { i } + 4 t \mathbf { j } + 3 t \mathbf { k }$, where $t$ is a constant. The line $l$ also passes through $B$.
Find the value of $t$.
\item The line $m$ has vector equation $\mathbf { r } = 5 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } + \mu ( a \mathbf { i } - \mathbf { j } + 3 \mathbf { k } )$. The acute angle between the directions of $l$ and $m$ is $\theta$, where $\cos \theta = \frac { 1 } { \sqrt { 6 } }$.\\
Find the possible values of $a$.\\
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-18_542_559_251_753}
A large cylindrical tank is used to store water. The base of the tank is a circle of radius 4 metres. At time $t$ minutes, the depth of the water in the tank is $h$ metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank.\\
(a) Show that $\frac { \mathrm { d } h } { \mathrm {~d} t } = - \lambda \sqrt { h }$, where $\lambda$ is a positive constant.\\
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-18_2718_42_107_2007}\\
(b) At time $t = 0$ the tap is opened. It is given that $h = 4$ when $t = 0$ and that $h = 2.25$ when $t = 20$.
Solve the differential equation to obtain an expression for $t$ in terms of $h$, and hence find the time taken to empty the tank.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q9 [10]}}