| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Standard +0.3 This is a structured multi-part question on 3D vectors with a real-world context. Parts (a)-(c) involve routine techniques: solving a quadratic for ground impact, substituting into a direction vector, and using scalar product for angles. Parts (d)-(e) require plane intersection and height comparison. Part (f) adds critical thinking about model limitations. While lengthy (6 parts), each step follows standard procedures with clear guidance, making it slightly easier than average despite the extended nature. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need k component to be zero at ground: \(0.84 + 0.8\lambda - \lambda^2 = 0 \Rightarrow \lambda = \ldots\) | M1 | Attempts to solve the quadratic from equating k component to zero |
| \(\lambda = -\frac{3}{5}, \frac{7}{5}\), but \(\lambda \geq 0\) so \(\lambda = \frac{7}{5}\) | A1 | Correct value, must select positive root, accept 1.4 or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction is \((9 - 4.6 \times 1.4)\mathbf{i} + 15\mathbf{j} + (0.8 - 2 \times 1.4)\) \(= 2.56\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}\) or \(\frac{64}{25}\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}\) | B1ft | For \((2.56, 15, -2)\) or follow through \((9 - 4.6\times\lambda', 15, 0.8 - 2\times\lambda')\) for their \(\lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction perpendicular to ground is \(a\mathbf{k}\); angle given by \(\cos\theta = \frac{a\mathbf{k}\cdot(2.56\mathbf{i}+15\mathbf{j}-2\mathbf{k})}{a \times | 2.56\mathbf{i}+15\mathbf{j}-2\mathbf{k} | }\) |
| \(= \frac{-2}{\sqrt{2.56^2 + 15^2 + (-2)^2}} (= -0.130\ldots)\) OR \(= \frac{231.5536}{\sqrt{2.56^2+15^2+(-2)^2}\sqrt{2.56^2+15^2+(0)^2}} = 0.991\ldots\) | M1 | Applies dot product formula \(\frac{a \cdot b}{ |
| \(90° - \arccos(-0.130\ldots) = -7.48\ldots\) or \(\arccos(0.991\ldots)\) | ddM1 | Dependent on both previous marks; correct method to proceed to required angle |
| So the tennis ball hits ground at angle of \(7.5°\) (1 d.p.) | A1 | cao |
| Alternative: Finds length of vector in ij plane \(= \sqrt{2.56^2 + 15^2}\) | M1 | Finds length of vector in ij plane |
| \(\tan\theta = \frac{2}{\sqrt{2.56^2 + 15^2}}\) | M1 | Finds the tan of the angle |
| \(\theta = \arctan\left(\frac{2}{\sqrt{2.56^2+15^2}}\right)\) or \(\theta = 90 - \arctan\left(\frac{\sqrt{2.56^2+15^2}}{2}\right)\) | ddM1 | Dependent on both previous marks; finds the required angle |
| \(7.5°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| In same plane as net when \(\mathbf{r}\cdot\mathbf{j} = 0\): \(-10.25 + 15\lambda = 0 \Rightarrow \lambda = \ldots \left(= \frac{41}{60}\right)\) | M1 | Attempts to find value of \(\lambda\) that gives zero j component |
| Position: \(\left(-4.1 + 9\times\frac{41}{60} - 2.3\left(\frac{41}{60}\right)^2\right)\mathbf{i} + 0\mathbf{j} + \left(0.84 + 0.8\times\frac{41}{60} - \left(\frac{41}{60}\right)^2\right)\mathbf{k}\) | M1 | Uses their value of \(\lambda\) in the equation of the path to find position |
| \(= \text{awrt } 0.976\mathbf{i} + \text{awrt } 0.920\mathbf{k}\) or \(= \text{awrt } 0.976\mathbf{i} + \frac{3311}{3600}\mathbf{k}\) | A1 | Correct position |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Modelling as a line, height of net is 0.9m along its length so as \(\mathbf{0.92 > 0.9}\) the ball will pass over the net according to the model. | B1ft | States \(0.920 > 0.9\) so according to the model the ball will pass over the net; follow through on their k component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies a suitable feature of the model that affects the outcome | M1 | Must reference the model; e.g. ball is not a particle / net is not a straight line. Do not accept wind/air resistance |
| Uses it to draw a compatible conclusion. Examples: ball is not a particle and will have diameter/radius; ball will clip the net; model says ball clears by 2cm which may be smaller than ball's diameter; net will not be straight/taut so will not be 0.9m high | A1 | Reasonable conclusion based on their reference to the model |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need **k** component to be zero at ground: $0.84 + 0.8\lambda - \lambda^2 = 0 \Rightarrow \lambda = \ldots$ | M1 | Attempts to solve the quadratic from equating **k** component to zero |
| $\lambda = -\frac{3}{5}, \frac{7}{5}$, but $\lambda \geq 0$ so $\lambda = \frac{7}{5}$ | A1 | Correct value, must select positive root, accept 1.4 or equivalent |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction is $(9 - 4.6 \times 1.4)\mathbf{i} + 15\mathbf{j} + (0.8 - 2 \times 1.4)$ $= 2.56\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}$ or $\frac{64}{25}\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}$ | B1ft | For $(2.56, 15, -2)$ or follow through $(9 - 4.6\times\lambda', 15, 0.8 - 2\times\lambda')$ for their $\lambda$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction perpendicular to ground is $a\mathbf{k}$; angle given by $\cos\theta = \frac{a\mathbf{k}\cdot(2.56\mathbf{i}+15\mathbf{j}-2\mathbf{k})}{a \times |2.56\mathbf{i}+15\mathbf{j}-2\mathbf{k}|}$ | M1 | Recognises angle between perpendicular and direction vector is needed; identifies perpendicular as $a\mathbf{k}$ for any non-zero $a$; attempts dot product |
| $= \frac{-2}{\sqrt{2.56^2 + 15^2 + (-2)^2}} (= -0.130\ldots)$ OR $= \frac{231.5536}{\sqrt{2.56^2+15^2+(-2)^2}\sqrt{2.56^2+15^2+(0)^2}} = 0.991\ldots$ | M1 | Applies dot product formula $\frac{a \cdot b}{|a||b|}$ correctly between any two vectors |
| $90° - \arccos(-0.130\ldots) = -7.48\ldots$ or $\arccos(0.991\ldots)$ | ddM1 | Dependent on both previous marks; correct method to proceed to required angle |
| So the tennis ball hits ground at angle of $7.5°$ (1 d.p.) | A1 | cao |
| **Alternative:** Finds length of vector in **ij** plane $= \sqrt{2.56^2 + 15^2}$ | M1 | Finds length of vector in **ij** plane |
| $\tan\theta = \frac{2}{\sqrt{2.56^2 + 15^2}}$ | M1 | Finds the tan of the angle |
| $\theta = \arctan\left(\frac{2}{\sqrt{2.56^2+15^2}}\right)$ or $\theta = 90 - \arctan\left(\frac{\sqrt{2.56^2+15^2}}{2}\right)$ | ddM1 | Dependent on both previous marks; finds the required angle |
| $7.5°$ | A1 | cao |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| In same plane as net when $\mathbf{r}\cdot\mathbf{j} = 0$: $-10.25 + 15\lambda = 0 \Rightarrow \lambda = \ldots \left(= \frac{41}{60}\right)$ | M1 | Attempts to find value of $\lambda$ that gives zero **j** component |
| Position: $\left(-4.1 + 9\times\frac{41}{60} - 2.3\left(\frac{41}{60}\right)^2\right)\mathbf{i} + 0\mathbf{j} + \left(0.84 + 0.8\times\frac{41}{60} - \left(\frac{41}{60}\right)^2\right)\mathbf{k}$ | M1 | Uses their value of $\lambda$ in the equation of the path to find position |
| $= \text{awrt } 0.976\mathbf{i} + \text{awrt } 0.920\mathbf{k}$ or $= \text{awrt } 0.976\mathbf{i} + \frac{3311}{3600}\mathbf{k}$ | A1 | Correct position |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Modelling as a line, height of net is 0.9m along its length so as $\mathbf{0.92 > 0.9}$ the ball **will** pass over the net according to the model. | B1ft | States $0.920 > 0.9$ so according to the model the ball will pass over the net; follow through on their **k** component |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies a suitable feature of the model that affects the outcome | M1 | Must reference the model; e.g. ball is not a particle / net is not a straight line. Do not accept wind/air resistance |
| Uses it to draw a compatible conclusion. Examples: ball is **not a particle** and will have **diameter/radius**; ball will **clip the net**; model says ball clears by 2cm which may be smaller than ball's **diameter**; net will **not be straight/taut** so will not be 0.9m high | A1 | Reasonable conclusion based on their reference to the model |
---\begin{enumerate}
\item The surface of a horizontal tennis court is modelled as part of a horizontal plane, with the origin on the ground at the centre of the court, and
\end{enumerate}
\begin{itemize}
\item i and j are unit vectors directed across the width and length of the court respectively
\item $\quad \mathbf { k }$ is a unit vector directed vertically upwards
\item units are metres
\end{itemize}
After being hit, a tennis ball, modelled as a particle, moves along the path with equation
$$\mathbf { r } = \left( - 4.1 + 9 \lambda - 2.3 \lambda ^ { 2 } \right) \mathbf { i } + ( - 10.25 + 15 \lambda ) \mathbf { j } + \left( 0.84 + 0.8 \lambda - \lambda ^ { 2 } \right) \mathbf { k }$$
where $\lambda$ is a scalar parameter with $\lambda \geqslant 0$\\
Assuming that the tennis ball continues on this path until it hits the ground,\\
(a) find the value of $\lambda$ at the point where the ball hits the ground.
The direction in which the tennis ball is moving at a general point on its path is given by
$$( 9 - 4.6 \lambda ) \mathbf { i } + 15 \mathbf { j } + ( 0.8 - 2 \lambda ) \mathbf { k }$$
(b) Write down the direction in which the tennis ball is moving as it hits the ground.\\
(c) Hence find the acute angle at which the tennis ball hits the ground, giving your answer in degrees to one decimal place.
The net of the tennis court lies in the plane $\mathbf { r } . \mathbf { j } = 0$\\
(d) Find the position of the tennis ball at the point where it is in the same plane as the net.
The maximum height above the court of the top of the net is 0.9 m .\\
Modelling the top of the net as a horizontal straight line,\\
(e) state whether the tennis ball will pass over the net according to the model, giving a reason for your answer.
With reference to the model,\\
(f) decide whether the tennis ball will actually pass over the net, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel CP AS 2022 Q6 [13]}}