| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring routine techniques: finding a line equation from two points, calculating angle between lines using dot product, finding a point on a line satisfying a perpendicularity condition, and computing triangle area. While it has multiple parts (8+ marks total), each step follows textbook methods without requiring novel insight or complex problem-solving, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\overrightarrow{AB} = \pm\left(\begin{pmatrix}5\\2\\7\end{pmatrix} - \begin{pmatrix}2\\1\\9\end{pmatrix}\right)\) | M1 | Correct attempt at direction. May be implied by at least 2 correct components if no method seen |
| \(\mathbf{r} = \begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix}\) or \(\mathbf{r} = 2\mathbf{i}+\mathbf{j}+9\mathbf{k}+\lambda(3\mathbf{i}+\mathbf{j}-2\mathbf{k})\) | A1 | Accept equivalents but must be equation and must be "\(\mathbf{r} =\)" or \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \ldots\) Do not allow bold unit vector notation without correct form seen earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}\) | M1 | Attempts \(\pm\overrightarrow{AC}\). May be implied by at least 2 correct components |
| \(\pm\overrightarrow{AB}\cdot\pm\overrightarrow{AC} = | AB | |
| \(\theta = 60°\) | A1 | cao (Must be degrees not radians) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB = \sqrt{14},\ AC = 2\sqrt{14},\ BC = \sqrt{42}\) | M1 | Attempts lengths of all 3 sides |
| \(42 = 14 + 56 - 2\sqrt{14}\sqrt{56}\cos\theta \Rightarrow \cos\theta = \frac{28}{2\sqrt{14}\sqrt{56}} \Rightarrow \theta = \ldots\) | dM1 | Attempt cosine rule and proceeds to \(\theta\) |
| \(\theta = 60°\) | A1 | cao (Must be degrees not radians) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}\) | M1 | Attempts \(\pm\overrightarrow{AC}\) |
| \(\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-4&-6\\3&1&-2\end{vmatrix} = \begin{pmatrix}14\\-14\\14\end{pmatrix} \Rightarrow \sin\theta = \frac{\sqrt{14^2+14^2+14^2}}{\sqrt{2^2+4^2+6^2}\sqrt{3^2+1^2+2^2}}\) | M1 | Attempt vector product of \(\pm\overrightarrow{AB}\) or direction vector from (a) and \(\pm\overrightarrow{AC}\), proceeds to \(\theta\) |
| \(\theta = 60°\) | A1 | cao (Must be degrees not radians) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix} - \begin{pmatrix}4\\-3\\3\end{pmatrix}\) | M1 | Attempts \(\overrightarrow{CD}\) by finding (general point on \(AB\)) \(- \overrightarrow{OC}\) or (their part (a)) \(- \overrightarrow{OC}\) |
| \(\begin{pmatrix}2\\-4\\-6\end{pmatrix}\cdot\begin{pmatrix}3\lambda-2\\\lambda+4\\-2\lambda+6\end{pmatrix} = 0\); \(6\lambda - 4 - 4\lambda - 16 - 36 + 12\lambda = 0 \Rightarrow \lambda = \ldots\) | M1 | Attempts \(\overrightarrow{AC}\cdot\overrightarrow{CD} = 0\) and solves for \(\lambda\). Must use correct \(AC\) or attempt at \(AC\), and correct attempt at \(CD\) |
| \(\lambda = \)"\(4\)"\(\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + \)"\(4\)"\(\begin{pmatrix}3\\1\\-2\end{pmatrix}\) | ddM1 | Uses their \(\lambda\) to find \(D\). Dependent on both previous M's |
| \((14, 5, 1)\) or \(14\mathbf{i}+5\mathbf{j}+\mathbf{k}\) or \(\begin{pmatrix}14\\5\\1\end{pmatrix}\) | A1 | Correct coordinates or vector and no other points or vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AC = 2\sqrt{14} \Rightarrow AD = \frac{2\sqrt{14}}{\cos 60°} (= 4\sqrt{14})\) | M1 | Correct attempt at length of \(AD\) |
| \(AB = \sqrt{14} \Rightarrow AD = 4AB\) or \((3\lambda)^2 + \lambda^2 + (2\lambda)^2 = (4\sqrt{14})^2 \Rightarrow \lambda = \ldots\) | M1 | Uses ratio of \(AB\) to \(AD\) to find value for "\(\lambda\)" or uses length of \(AD\) and applies Pythagoras |
| \(\lambda = \)"\(4\)"\(\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + \)"\(4\)"\(\begin{pmatrix}3\\1\\-2\end{pmatrix}\) | ddM1 | Uses their "\(\lambda\)" to find \(D\). Dependent on both previous M's |
| \(D(14, 5, 1)\) or \(14\mathbf{i}+5\mathbf{j}+\mathbf{k}\) etc. | A1 | Correct coordinates or vector and no other points or vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix} - \begin{pmatrix}4\\-3\\3\end{pmatrix}\) | M1 | Attempts \(\overrightarrow{CD}\) by finding (general point on \(AB\)) \(- \overrightarrow{OC}\) or (their part (a)) \(- \overrightarrow{OC}\) |
| \((3\lambda-2)^2 + (\lambda+4)^2 + (6-2\lambda)^2 = (AC\tan 60°)^2\); \(\lambda^2 - 2\lambda - 8 = 0 \Rightarrow \lambda = \ldots\) | M1 | Attempts \((CD)^2 = (AC\tan 60°)^2\) and solves for \(\lambda\) |
| \(\lambda = \)"\(4\)"\(\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + \)"\(4\)"\(\begin{pmatrix}3\\1\\-2\end{pmatrix}\) | ddM1 | Uses their value of \(\lambda\) to find \(D\). Dependent on both previous M's |
| \((14, 5, 1)\) or \(14\mathbf{i}+5\mathbf{j}+\mathbf{k}\) or \(\begin{pmatrix}14\\5\\1\end{pmatrix}\) | A1 | Correct coordinates or vector and no other points or vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\1\\9\end{pmatrix}+\lambda\begin{pmatrix}3\\1\\-2\end{pmatrix}-\begin{pmatrix}4\\-3\\3\end{pmatrix}\) | M1 | Attempts \(\overrightarrow{CD}\) by finding (a general point on \(AB\)) \(- \overrightarrow{OC}\) or (their part (a)) \(- \overrightarrow{OC}\) |
| \((3\lambda-2)^2+(\lambda+4)^2+(6-2\lambda)^2+AC^2=(3\lambda)^2+\lambda^2+(2\lambda)^2\); \(28\lambda-112=0 \Rightarrow \lambda=\ldots\) Attempts \(AC^2+CD^2=AD^2\) and solves for \(\lambda\) | M1 | |
| \(\lambda=4 \Rightarrow OD=\begin{pmatrix}2\\1\\9\end{pmatrix}+4\begin{pmatrix}3\\1\\-2\end{pmatrix}\) | ddM1 | Uses their value of \(\lambda\) to find \(D\). Dependent on both previous M's |
| \((14,5,1)\) or \(14\mathbf{i}+5\mathbf{j}+\mathbf{k}\) or \(\begin{pmatrix}14\\5\\1\end{pmatrix}\) | A1 | Correct coordinates or vector and no other points or vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(ADC = \frac{1}{2}AC\times CD = \frac{1}{2}\sqrt{56}\sqrt{168}\) | M1 | Correct triangle area method |
| \(=28\sqrt{3}\) | A1 | cao |
| Alternatives: \(\frac{1}{2}AC\times AD\sin 60° = \frac{1}{2}\sqrt{56}\sqrt{224}\cdot\frac{\sqrt{3}}{2}\); \(\frac{1}{2}AD\times DC\sin 30°=\frac{1}{2}\sqrt{168}\sqrt{224}\cdot\frac{1}{2}\); \(\frac{1}{2}AC\times AC\tan 60°=\frac{1}{2}\sqrt{56}\sqrt{56}\sqrt{3}\) |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AB} = \pm\left(\begin{pmatrix}5\\2\\7\end{pmatrix} - \begin{pmatrix}2\\1\\9\end{pmatrix}\right)$ | M1 | Correct attempt at direction. May be implied by at least 2 correct components if no method seen |
| $\mathbf{r} = \begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix}$ or $\mathbf{r} = 2\mathbf{i}+\mathbf{j}+9\mathbf{k}+\lambda(3\mathbf{i}+\mathbf{j}-2\mathbf{k})$ | A1 | Accept equivalents but must be equation and must be "$\mathbf{r} =$" or $\begin{pmatrix}x\\y\\z\end{pmatrix} = \ldots$ Do not allow bold unit vector notation without correct form seen earlier |
**(2 marks)**
---
## Question 6(b):
### Way 1 (Scalar Product):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}$ | M1 | Attempts $\pm\overrightarrow{AC}$. May be implied by at least 2 correct components |
| $\pm\overrightarrow{AB}\cdot\pm\overrightarrow{AC} = |AB||AC|\cos\theta \Rightarrow 6-4+12 = \sqrt{14}\sqrt{56}\cos\theta \Rightarrow \cos\theta = \frac{14}{\sqrt{14}\sqrt{56}} \Rightarrow \theta = \ldots$ | dM1 | Attempt scalar product of $\pm\overrightarrow{AB}$ or direction vector from (a) and $\pm\overrightarrow{AC}$, proceeds to $\theta$ |
| $\theta = 60°$ | A1 | cao (Must be **degrees** not radians) |
**(3 marks)**
### Way 2 (Cosine Rule):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = \sqrt{14},\ AC = 2\sqrt{14},\ BC = \sqrt{42}$ | M1 | Attempts lengths of all 3 sides |
| $42 = 14 + 56 - 2\sqrt{14}\sqrt{56}\cos\theta \Rightarrow \cos\theta = \frac{28}{2\sqrt{14}\sqrt{56}} \Rightarrow \theta = \ldots$ | dM1 | Attempt cosine rule and proceeds to $\theta$ |
| $\theta = 60°$ | A1 | cao (Must be **degrees** not radians) |
**(3 marks)**
### Way 3 (Vector Product):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}$ | M1 | Attempts $\pm\overrightarrow{AC}$ |
| $\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-4&-6\\3&1&-2\end{vmatrix} = \begin{pmatrix}14\\-14\\14\end{pmatrix} \Rightarrow \sin\theta = \frac{\sqrt{14^2+14^2+14^2}}{\sqrt{2^2+4^2+6^2}\sqrt{3^2+1^2+2^2}}$ | M1 | Attempt vector product of $\pm\overrightarrow{AB}$ or direction vector from (a) and $\pm\overrightarrow{AC}$, proceeds to $\theta$ |
| $\theta = 60°$ | A1 | cao (Must be **degrees** not radians) |
**(3 marks)**
---
## Question 6(c):
### Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix} - \begin{pmatrix}4\\-3\\3\end{pmatrix}$ | M1 | Attempts $\overrightarrow{CD}$ by finding (general point on $AB$) $- \overrightarrow{OC}$ or (their part (a)) $- \overrightarrow{OC}$ |
| $\begin{pmatrix}2\\-4\\-6\end{pmatrix}\cdot\begin{pmatrix}3\lambda-2\\\lambda+4\\-2\lambda+6\end{pmatrix} = 0$; $6\lambda - 4 - 4\lambda - 16 - 36 + 12\lambda = 0 \Rightarrow \lambda = \ldots$ | M1 | Attempts $\overrightarrow{AC}\cdot\overrightarrow{CD} = 0$ and solves for $\lambda$. Must use correct $AC$ or attempt at $AC$, and correct attempt at $CD$ |
| $\lambda = $"$4$"$\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + $"$4$"$\begin{pmatrix}3\\1\\-2\end{pmatrix}$ | ddM1 | Uses their $\lambda$ to find $D$. Dependent on both previous M's |
| $(14, 5, 1)$ or $14\mathbf{i}+5\mathbf{j}+\mathbf{k}$ or $\begin{pmatrix}14\\5\\1\end{pmatrix}$ | A1 | Correct coordinates or vector and no other points or vectors |
**(4 marks)**
### Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AC = 2\sqrt{14} \Rightarrow AD = \frac{2\sqrt{14}}{\cos 60°} (= 4\sqrt{14})$ | M1 | Correct attempt at length of $AD$ |
| $AB = \sqrt{14} \Rightarrow AD = 4AB$ or $(3\lambda)^2 + \lambda^2 + (2\lambda)^2 = (4\sqrt{14})^2 \Rightarrow \lambda = \ldots$ | M1 | Uses ratio of $AB$ to $AD$ to find value for "$\lambda$" or uses length of $AD$ and applies Pythagoras |
| $\lambda = $"$4$"$\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + $"$4$"$\begin{pmatrix}3\\1\\-2\end{pmatrix}$ | ddM1 | Uses their "$\lambda$" to find $D$. Dependent on both previous M's |
| $D(14, 5, 1)$ or $14\mathbf{i}+5\mathbf{j}+\mathbf{k}$ etc. | A1 | Correct coordinates or vector and **no other points or vectors** |
### Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\1\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix} - \begin{pmatrix}4\\-3\\3\end{pmatrix}$ | M1 | Attempts $\overrightarrow{CD}$ by finding (general point on $AB$) $- \overrightarrow{OC}$ or (their part (a)) $- \overrightarrow{OC}$ |
| $(3\lambda-2)^2 + (\lambda+4)^2 + (6-2\lambda)^2 = (AC\tan 60°)^2$; $\lambda^2 - 2\lambda - 8 = 0 \Rightarrow \lambda = \ldots$ | M1 | Attempts $(CD)^2 = (AC\tan 60°)^2$ and solves for $\lambda$ |
| $\lambda = $"$4$"$\Rightarrow OD = \begin{pmatrix}2\\1\\9\end{pmatrix} + $"$4$"$\begin{pmatrix}3\\1\\-2\end{pmatrix}$ | ddM1 | Uses their value of $\lambda$ to find $D$. Dependent on both previous M's |
| $(14, 5, 1)$ or $14\mathbf{i}+5\mathbf{j}+\mathbf{k}$ or $\begin{pmatrix}14\\5\\1\end{pmatrix}$ | A1 | Correct coordinates or vector and no other points or vectors |
# Question 6 (Vectors):
## Part (c) Way 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\1\\9\end{pmatrix}+\lambda\begin{pmatrix}3\\1\\-2\end{pmatrix}-\begin{pmatrix}4\\-3\\3\end{pmatrix}$ | M1 | Attempts $\overrightarrow{CD}$ by finding (a general point on $AB$) $- \overrightarrow{OC}$ or (their part (a)) $- \overrightarrow{OC}$ |
| $(3\lambda-2)^2+(\lambda+4)^2+(6-2\lambda)^2+AC^2=(3\lambda)^2+\lambda^2+(2\lambda)^2$; $28\lambda-112=0 \Rightarrow \lambda=\ldots$ Attempts $AC^2+CD^2=AD^2$ and solves for $\lambda$ | M1 | |
| $\lambda=4 \Rightarrow OD=\begin{pmatrix}2\\1\\9\end{pmatrix}+4\begin{pmatrix}3\\1\\-2\end{pmatrix}$ | ddM1 | Uses their value of $\lambda$ to find $D$. **Dependent on both previous M's** |
| $(14,5,1)$ or $14\mathbf{i}+5\mathbf{j}+\mathbf{k}$ or $\begin{pmatrix}14\\5\\1\end{pmatrix}$ | A1 | Correct coordinates or vector and no other points or vectors |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $ADC = \frac{1}{2}AC\times CD = \frac{1}{2}\sqrt{56}\sqrt{168}$ | M1 | Correct triangle area method |
| $=28\sqrt{3}$ | A1 | cao |
| Alternatives: $\frac{1}{2}AC\times AD\sin 60° = \frac{1}{2}\sqrt{56}\sqrt{224}\cdot\frac{\sqrt{3}}{2}$; $\frac{1}{2}AD\times DC\sin 30°=\frac{1}{2}\sqrt{168}\sqrt{224}\cdot\frac{1}{2}$; $\frac{1}{2}AC\times AC\tan 60°=\frac{1}{2}\sqrt{56}\sqrt{56}\sqrt{3}$ | | |
---6. Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have coordinates ( $2,1,9 ) , ( 5,2,7 )$ and $( 4 , - 3,3 )$ respectively.
The line $l$ passes through the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for the line $l$.
\item Find, in degrees, the acute angle between the line $I$ and the line $A C$.
The point $D$ lies on the line $l$ such that angle $A C D$ is $90 ^ { \circ }$
\item Find the coordinates of $D$.
\item Find the exact area of triangle $A D C$, giving your answer as a fully simplified surd.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q6 [11]}}