| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.3 This is a standard two-part vectors question requiring (a) using the scalar product formula with the angle between lines (routine application of cos θ = |a·b|/(|a||b|)), and (b) finding perpendicular distance using the standard formula or vector projection method. Both parts are textbook exercises with straightforward execution, making it slightly easier than average for A-level Further Maths. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the correct process to calculate the scalar product of the direction vectors | M1 | \((-2 + 4 + 2c)\) |
| Divide the scalar product by the product of the moduli and equate the result to \(\cos 60°\) | M1 | Or equivalent e.g. \(2 + 2c = \sqrt{6}\sqrt{20+c^2}\cos 60°\). Allow for the correct process using \(60°\) but the wrong vectors. |
| Obtain correct equation in \(c\) | A1 | e.g. \(\frac{2+2c}{\sqrt{6}\sqrt{20+c^2}} = \frac{1}{2}\) or \(10c^2 + 32c - 104 = 0\) |
| Obtain \(c = 2\) | A1 | Only. |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calling \((6, -3, 6)\) \(A\), find \(\overrightarrow{AP}\) for a general point \(P\) on \(l\) | B1 | e.g. \(\begin{pmatrix}-3+\lambda \\ 1+\lambda \\ -5+2\lambda\end{pmatrix}\) |
| Equate the scalar product of *their* \(\overrightarrow{AP}\) and a direction vector for \(l\) to zero and obtain an equation in \(\lambda\) | \*M1 | e.g. \((-3+\lambda)+(1+\lambda)+(-10+4\lambda)=0\) |
| Solve and obtain \(\lambda = 2\) | A1 | |
| Carry out a method to calculate \( | \overrightarrow{AP} | \) |
| Obtain \(\sqrt{11}\) from correct working | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calling \((6, -3, 6)\) \(A\), find \(\overrightarrow{AP}\) for a general point \(P\) on \(l\) | B1 | e.g. \(\begin{pmatrix}-3+\lambda \\ 1+\lambda \\ -5+2\lambda\end{pmatrix}\) |
| Differentiate the modulus of \(\overrightarrow{AP}\) or the square of the modulus and equate the derivative to zero | \*M1 | e.g. \(2(-3+\lambda)+2(1+\lambda)+4(-5+2\lambda)=0\) |
| Solve and obtain \(\lambda = 2\) | A1 | |
| Carry out a method to calculate \( | \overrightarrow{AP} | \) |
| Obtain \(\sqrt{11}\) from correct working | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vector from \((6,-3,6)\) to \((3,-2,1)\) is \(-3\mathbf{i}+\mathbf{j}-5\mathbf{k}\) | B1 | The method works for vector from \((6,-3,6)\) to any point on \(l\) |
| Use scalar product to find the angle between *their* vector and the direction of \(l\) | M1 | |
| Obtain \(\cos\theta = \dfrac{3-1+10}{\sqrt{35}\sqrt{6}}\left(=\sqrt{\dfrac{24}{35}}\right)\) or \(\sin\theta=\sqrt{\dfrac{11}{35}}\) | A1 | |
| Correct use of trig to find the projection of their vector on the normal to \(l\) | M1 | \(\sqrt{35}\sin\theta = \sqrt{35}\times\sqrt{\dfrac{11}{35}}\) |
| Obtain \(\sqrt{11}\) from correct working | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vector from \((6,-3,6)\) to \((3,-2,1)\) is \(-3\mathbf{i}+\mathbf{j}-5\mathbf{k}\) | B1 | |
| Find the vector product of *their* vector and the direction of \(l\) | M1 | |
| Obtain \(\mathbf{i}(2+5)-\mathbf{j}(-6+5)+\mathbf{k}(-3-1)(=7\mathbf{i}+\mathbf{j}-4\mathbf{k})\) | A1 | |
| Correct use of trig to find the perpendicular distance | M1 | \(\dfrac{ |
| Distance \(=\dfrac{\sqrt{66}}{\sqrt{6}}=\sqrt{11}\) | A1 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the correct process to calculate the scalar product of the direction vectors | M1 | $(-2 + 4 + 2c)$ |
| Divide the scalar product by the product of the moduli and equate the result to $\cos 60°$ | M1 | Or equivalent e.g. $2 + 2c = \sqrt{6}\sqrt{20+c^2}\cos 60°$. Allow for the correct process using $60°$ but the wrong vectors. |
| Obtain correct equation in $c$ | A1 | e.g. $\frac{2+2c}{\sqrt{6}\sqrt{20+c^2}} = \frac{1}{2}$ or $10c^2 + 32c - 104 = 0$ |
| Obtain $c = 2$ | A1 | Only. |
| **Total** | **4** | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calling $(6, -3, 6)$ $A$, find $\overrightarrow{AP}$ for a general point $P$ on $l$ | B1 | e.g. $\begin{pmatrix}-3+\lambda \\ 1+\lambda \\ -5+2\lambda\end{pmatrix}$ |
| Equate the scalar product of *their* $\overrightarrow{AP}$ and a direction vector for $l$ to zero and obtain an equation in $\lambda$ | \*M1 | e.g. $(-3+\lambda)+(1+\lambda)+(-10+4\lambda)=0$ |
| Solve and obtain $\lambda = 2$ | A1 | |
| Carry out a method to calculate $|\overrightarrow{AP}|$ | DM1 | e.g. $(-1)^2+3^2+(-1)^2$ or $1^2+3^2+1^2$ |
| Obtain $\sqrt{11}$ from correct working | A1 | AG |
**Alternative method 1 for question 10(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Calling $(6, -3, 6)$ $A$, find $\overrightarrow{AP}$ for a general point $P$ on $l$ | B1 | e.g. $\begin{pmatrix}-3+\lambda \\ 1+\lambda \\ -5+2\lambda\end{pmatrix}$ |
| Differentiate the modulus of $\overrightarrow{AP}$ or the square of the modulus and equate the derivative to zero | \*M1 | e.g. $2(-3+\lambda)+2(1+\lambda)+4(-5+2\lambda)=0$ |
| Solve and obtain $\lambda = 2$ | A1 | |
| Carry out a method to calculate $|\overrightarrow{AP}|$ | DM1 | e.g. $(-1)^2+3^2+(-1)^2$ or $1^2+3^2+1^2$ |
| Obtain $\sqrt{11}$ from correct working | A1 | AG |
## Question 10(b) – Alternative Method 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vector from $(6,-3,6)$ to $(3,-2,1)$ is $-3\mathbf{i}+\mathbf{j}-5\mathbf{k}$ | B1 | The method works for vector from $(6,-3,6)$ to any point on $l$ |
| Use scalar product to find the angle between *their* vector and the direction of $l$ | M1 | |
| Obtain $\cos\theta = \dfrac{3-1+10}{\sqrt{35}\sqrt{6}}\left(=\sqrt{\dfrac{24}{35}}\right)$ or $\sin\theta=\sqrt{\dfrac{11}{35}}$ | A1 | |
| Correct use of trig to find the projection of their vector on the normal to $l$ | M1 | $\sqrt{35}\sin\theta = \sqrt{35}\times\sqrt{\dfrac{11}{35}}$ |
| Obtain $\sqrt{11}$ from correct working | A1 | AG |
## Question 10(c) – Alternative Method 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vector from $(6,-3,6)$ to $(3,-2,1)$ is $-3\mathbf{i}+\mathbf{j}-5\mathbf{k}$ | B1 | |
| Find the vector product of *their* vector and the direction of $l$ | M1 | |
| Obtain $\mathbf{i}(2+5)-\mathbf{j}(-6+5)+\mathbf{k}(-3-1)(=7\mathbf{i}+\mathbf{j}-4\mathbf{k})$ | A1 | |
| Correct use of trig to find the perpendicular distance | M1 | $\dfrac{|\text{their vector product}|}{|\text{direction vector}|}$ |
| Distance $=\dfrac{\sqrt{66}}{\sqrt{6}}=\sqrt{11}$ | A1 | |
**Total: 5 marks**
---10 The equations of the lines $l$ and $m$ are given by
$$l : \mathbf { r } = \left( \begin{array} { r }
3 \\
- 2 \\
1
\end{array} \right) + \lambda \left( \begin{array} { l }
1 \\
1 \\
2
\end{array} \right) \quad \text { and } \quad m : \mathbf { r } = \left( \begin{array} { r }
6 \\
- 3 \\
6
\end{array} \right) + \mu \left( \begin{array} { r }
- 2 \\
4 \\
c
\end{array} \right)$$
where $c$ is a positive constant. It is given that the angle between $l$ and $m$ is $60 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $c$.
\item Show that the length of the perpendicular from $( 6 , - 3,6 )$ to $l$ is $\sqrt { 11 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q10 [9]}}