| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: finding a normal vector via cross product for the Cartesian equation, substituting parametric line equations into the plane, and using the scalar product formula for angle between planes. All methods are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts normal vector: e.g. let \(\mathbf{n}=a\mathbf{i}+b\mathbf{j}+\mathbf{k}\), then \(a+2b-3=0,\ -a+2b+1=0 \Rightarrow a=\ldots, b=\ldots\) or \(\mathbf{n}=(\mathbf{i}+2\mathbf{j}-3\mathbf{k})\times(-\mathbf{i}+2\mathbf{j}+\mathbf{k})\) | M1 | |
| \(\mathbf{n} = k(4\mathbf{i}+\mathbf{j}+2\mathbf{k})\) | A1 | |
| \((4\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}+4\mathbf{j}-\mathbf{k}) = \ldots\) | M1 | |
| \(4x+y+2z=10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=2+\lambda-\mu,\ y=4+2\lambda+2\mu,\ z=-1-3\lambda+\mu \Rightarrow 2x+y=8+4\lambda,\ y-2z=6+8\lambda\) | M1, A1 | |
| \(2(2x+y-8)=y-2z-6 \Rightarrow (4x+y+2z=10)\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{x-1}{5}=\frac{y-3}{-3}=\frac{z+2}{4} \Rightarrow \mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}-3\mathbf{j}+4\mathbf{k})\); \(4(1+5\lambda)+3-3\lambda+2(4\lambda-2)=10 \Rightarrow \lambda=\ldots\) | M1 | |
| \(\lambda=\frac{7}{25} \Rightarrow \mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\frac{7}{25}(5\mathbf{i}-3\mathbf{j}+4\mathbf{k})\) | dM1 | |
| \(\left(\frac{12}{5},\ \frac{54}{25},\ -\frac{22}{25}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4x+\left(-\frac{3}{5}(x-1)+3\right)+2\left(\frac{4}{5}(x-1)-2\right)=10 \Rightarrow x=\ldots\) | M1 | |
| \(\Rightarrow y=\ldots,\ z=\ldots\) | M1 | |
| \(\left(\frac{12}{5},\ \frac{54}{25},\ -\frac{22}{25}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((4\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}-\mathbf{j}+3\mathbf{k})=8-1+6=13\); \(13=\sqrt{14}\sqrt{21}\cos\theta \Rightarrow \theta=\ldots\) | M1 | |
| \(\theta = 41°\) | A1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts normal vector: e.g. let $\mathbf{n}=a\mathbf{i}+b\mathbf{j}+\mathbf{k}$, then $a+2b-3=0,\ -a+2b+1=0 \Rightarrow a=\ldots, b=\ldots$ or $\mathbf{n}=(\mathbf{i}+2\mathbf{j}-3\mathbf{k})\times(-\mathbf{i}+2\mathbf{j}+\mathbf{k})$ | M1 | |
| $\mathbf{n} = k(4\mathbf{i}+\mathbf{j}+2\mathbf{k})$ | A1 | |
| $(4\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}+4\mathbf{j}-\mathbf{k}) = \ldots$ | M1 | |
| $4x+y+2z=10$ | A1 | |
### Alternative 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=2+\lambda-\mu,\ y=4+2\lambda+2\mu,\ z=-1-3\lambda+\mu \Rightarrow 2x+y=8+4\lambda,\ y-2z=6+8\lambda$ | M1, A1 | |
| $2(2x+y-8)=y-2z-6 \Rightarrow (4x+y+2z=10)$ | M1, A1 | |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x-1}{5}=\frac{y-3}{-3}=\frac{z+2}{4} \Rightarrow \mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}-3\mathbf{j}+4\mathbf{k})$; $4(1+5\lambda)+3-3\lambda+2(4\lambda-2)=10 \Rightarrow \lambda=\ldots$ | M1 | |
| $\lambda=\frac{7}{25} \Rightarrow \mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\frac{7}{25}(5\mathbf{i}-3\mathbf{j}+4\mathbf{k})$ | dM1 | |
| $\left(\frac{12}{5},\ \frac{54}{25},\ -\frac{22}{25}\right)$ | A1 | |
### Alternative 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4x+\left(-\frac{3}{5}(x-1)+3\right)+2\left(\frac{4}{5}(x-1)-2\right)=10 \Rightarrow x=\ldots$ | M1 | |
| $\Rightarrow y=\ldots,\ z=\ldots$ | M1 | |
| $\left(\frac{12}{5},\ \frac{54}{25},\ -\frac{22}{25}\right)$ | A1 | |
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4\mathbf{i}+\mathbf{j}+2\mathbf{k})\cdot(2\mathbf{i}-\mathbf{j}+3\mathbf{k})=8-1+6=13$; $13=\sqrt{14}\sqrt{21}\cos\theta \Rightarrow \theta=\ldots$ | M1 | |
| $\theta = 41°$ | A1 | |\begin{enumerate}
\item The plane $\Pi _ { 1 }$ has equation
\end{enumerate}
$$\mathbf { r } = 2 \mathbf { i } + 4 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k } ) + \mu ( - \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$$
where $\lambda$ and $\mu$ are scalar parameters.\\
(a) Find a Cartesian equation for $\Pi _ { 1 }$
The line $l$ has equation
$$\frac { x - 1 } { 5 } = \frac { y - 3 } { - 3 } = \frac { z + 2 } { 4 }$$
(b) Find the coordinates of the point of intersection of $l$ with $\Pi _ { 1 }$
The plane $\Pi _ { 2 }$ has equation
$$\mathbf { r . } ( 2 \mathbf { i } - \mathbf { j } + 3 \mathbf { k } ) = 5$$
(c) Find, to the nearest degree, the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$
\hfill \mbox{\textit{Edexcel CP1 2020 Q4 [9]}}