| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a multi-part question on 3D vectors covering standard techniques: showing perpendicularity via dot product (routine), finding a plane equation from a line and normal vector (standard method), verification by substitution (trivial), and finding a point on a line at a given distance (straightforward application of distance formula). All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}3\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=3\{+0\}-3\) | M1 | Applies dot product to direction vectors, minimum requirement is \(3-3\) |
| \(=0\) therefore lines are perpendicular | A1 | Shows dot product \(=0\) and concludes perpendicular |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=\begin{pmatrix}-2\\2\\0\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=\ldots\{2\}\) | M1 | |
| \(x+2y-3z=2\) o.e. | A1 | Note: \(\mathbf{i}+2\mathbf{j}-3\mathbf{k}=2\) is M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3+2(1)-3(1)=2\) (therefore lies on the plane) | B1 | See scheme, no conclusion required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}p\\q\\r\end{pmatrix}=\begin{pmatrix}2\\3\\2\end{pmatrix}+\mu\begin{pmatrix}1\\2\\-3\end{pmatrix}\) giving \(p=2+\mu\), \(q=3+2\mu\), \(r=2-3\mu\) | M1 | Uses intersection point to find coordinates of \(B\) as functions of parameter |
| \((p-3)^2+(q-1)^2+(r-1)^2=(2\sqrt{5})^2\) leading to \((-1+\mu)^2+(2+2\mu)^2+(1-3\mu)^2=20\) | M1 | Uses distance between \(A\) and \(B\) to form equation in parameter only |
| \(14\mu^2-14=0\) o.e. | A1 | Correct simplified quadratic |
| Solves quadratic: \(\{\mu=-1\) or \(\mu=1\}\) | M1 | |
| Uses \(\mu=-1\) only (discarding \(\mu=1\) which gives point \(A\)) | ddM1 | |
| \((1,1,5)\) only | A1 | |
| Alternative: \( | AX | =\sqrt{(3-2)^2+(1-3)^2+(1-2)^2}=\sqrt{6}\) |
| \( | XB | =\sqrt{(2\sqrt{5})^2-6}=\sqrt{14}\) |
| Find magnitude of direction vector and compare to \( | XB | \) to find \(\mu\) |
| \(\mu=-1\) or \(\mu=1\) | A1 | |
| \((1,1,5)\) only | A1 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=3\{+0\}-3$ | M1 | Applies dot product to direction vectors, minimum requirement is $3-3$ |
| $=0$ therefore lines are **perpendicular** | A1 | Shows dot product $=0$ and concludes perpendicular |
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=\begin{pmatrix}-2\\2\\0\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-3\end{pmatrix}=\ldots\{2\}$ | M1 | |
| $x+2y-3z=2$ o.e. | A1 | Note: $\mathbf{i}+2\mathbf{j}-3\mathbf{k}=2$ is M1A0 |
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3+2(1)-3(1)=2$ (therefore lies on the plane) | B1 | See scheme, no conclusion required |
# Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}p\\q\\r\end{pmatrix}=\begin{pmatrix}2\\3\\2\end{pmatrix}+\mu\begin{pmatrix}1\\2\\-3\end{pmatrix}$ giving $p=2+\mu$, $q=3+2\mu$, $r=2-3\mu$ | M1 | Uses intersection point to find coordinates of $B$ as functions of parameter |
| $(p-3)^2+(q-1)^2+(r-1)^2=(2\sqrt{5})^2$ leading to $(-1+\mu)^2+(2+2\mu)^2+(1-3\mu)^2=20$ | M1 | Uses distance between $A$ and $B$ to form equation in parameter only |
| $14\mu^2-14=0$ o.e. | A1 | Correct simplified quadratic |
| Solves quadratic: $\{\mu=-1$ or $\mu=1\}$ | M1 | |
| Uses $\mu=-1$ only (discarding $\mu=1$ which gives point $A$) | ddM1 | |
| $(1,1,5)$ only | A1 | |
| **Alternative:** $|AX|=\sqrt{(3-2)^2+(1-3)^2+(1-2)^2}=\sqrt{6}$ | M1 | |
| $|XB|=\sqrt{(2\sqrt{5})^2-6}=\sqrt{14}$ | M1 | Correctly uses Pythagoras |
| Find magnitude of direction vector and compare to $|XB|$ to find $\mu$ | M1 | |
| $\mu=-1$ or $\mu=1$ | A1 | |
| $(1,1,5)$ only | A1 | |\begin{enumerate}
\item The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } - 2 \\ 2 \\ 0 \end{array} \right) + \lambda \left( \begin{array} { l } 3 \\ 0 \\ 1 \end{array} \right)$ where $\lambda$ is a scalar parameter.
\end{enumerate}
The line $l _ { 2 }$ is parallel to $\left( \begin{array} { r } 1 \\ 2 \\ - 3 \end{array} \right)$\\
(a) Show that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular.
The plane $\Pi$ contains the line $l _ { 1 }$ and is perpendicular to $\left( \begin{array} { r } 1 \\ 2 \\ - 3 \end{array} \right)$\\
(b) Determine a Cartesian equation of $\Pi$\\
(c) Verify that the point $A ( 3,1,1 )$ lies on $\Pi$
Given that
\begin{itemize}
\item the point of intersection of $\Pi$ and $l _ { 2 }$ has coordinates $( 2,3,2 )$
\item the point $B ( p , q , r )$ lies on $l _ { 2 }$
\item the distance $A B$ is $2 \sqrt { 5 }$
\item $p , q$ and $r$ are positive integers\\
(d) determine the coordinates of $B$.
\end{itemize}
\hfill \mbox{\textit{Edexcel CP AS 2023 Q6 [11]}}