Edexcel C4 2017 June — Question 6 13 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2017
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeArea of parallelogram or trapezium using vectors
DifficultyStandard +0.3 This is a standard multi-part vectors question covering routine techniques: finding intersection of lines, angle between lines using dot product, distance calculations, and perpendicular conditions. All parts follow textbook methods with no novel insight required. Part (e) adds mild complexity by requiring consideration of two cases, but the overall question is slightly easier than average due to its structured, methodical nature.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles
6. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 4 \\ 28 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } - 1 \\ - 5 \\ 1 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 5 \\ 3 \\ 1 \end{array} \right) + \mu \left( \begin{array} { r } 3 \\ 0 \\ - 4 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(X\).
  1. Find the coordinates of the point \(X\).
  2. Find the size of the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to 2 decimal places. The point \(A\) lies on \(l _ { 1 }\) and has position vector \(\left( \begin{array} { r } 2 \\ 18 \\ 6 \end{array} \right)\)
  3. Find the distance \(A X\), giving your answer as a surd in its simplest form. The point \(Y\) lies on \(l _ { 2 }\). Given that the vector \(\overrightarrow { Y A }\) is perpendicular to the line \(l _ { 1 }\)
  4. find the distance \(Y A\), giving your answer to one decimal place. The point \(B\) lies on \(l _ { 1 }\) where \(| \overrightarrow { A X } | = 2 | \overrightarrow { A B } |\).
  5. Find the two possible position vectors of \(B\).
Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\{l_1=l_2\Rightarrow\}\ 28-5\lambda=3\ \{\Rightarrow\lambda=5\}\) or \(4-\lambda=5+3\mu\) and \(4+\lambda=1-4\mu\ \{\Rightarrow\mu=-2\}\)B1 \(28-5\lambda=3\) or \(4-\lambda=5+3\mu\) and \(4+\lambda=1-4\mu\); or \(\lambda=5\) or \(\mu=-2\) (can be implied)
\(\left\{\overrightarrow{OX}=\right\}\begin{pmatrix}4\\28\\4\end{pmatrix}+5\begin{pmatrix}-1\\-5\\1\end{pmatrix}\) or \(\begin{pmatrix}5\\3\\1\end{pmatrix}-2\begin{pmatrix}3\\0\\-4\end{pmatrix}\)M1 Puts \(l_1=l_2\) and solves to find \(\lambda\) and/or \(\mu\); and substitutes their value for \(\lambda\) into \(l_1\) or their value for \(\mu\) into \(l_2\)
So, \(X(-1,3,9)\)A1 cao \((-1,3,9)\) or \(\begin{pmatrix}-1\\3\\9\end{pmatrix}\) or \(-\mathbf{i}+3\mathbf{j}+9\mathbf{k}\) or condone \(\begin{pmatrix}-1\\3\\9\end{pmatrix}\)
Part (b) Way 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{d}_1=\begin{pmatrix}-1\\-5\\1\end{pmatrix}\), \(\mathbf{d}_2=\begin{pmatrix}3\\0\\-4\end{pmatrix}\Rightarrow\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}\)M1 Realisation that the dot product is required between \(\mathbf{d}_1\) and \(\mathbf{d}_2\), or a multiple of \(\mathbf{d}_1\) and \(\mathbf{d}_2\)
\(\cos\theta=\frac{\pm\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}}{\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}}\left\{=\frac{-7}{\sqrt{27}\cdot\sqrt{25}}\right\}\)dM1 Dependent on 1st M mark; applies dot product formula between \(\mathbf{d}_1\) and \(\mathbf{d}_2\) or a multiple
\(\{\theta=105.6303588...\Rightarrow\}\ \theta_{\text{Acute}}=74.36964117...=74.37\ (2\text{ dp})\)A1 awrt 74.37 seen in (b) only
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{AX}=``\overrightarrow{OX}"-\overrightarrow{OA}=\begin{pmatrix}-1\\3\\9\end{pmatrix}-\begin{pmatrix}2\\18\\6\end{pmatrix}=\begin{pmatrix}-3\\-15\\3\end{pmatrix}\) or \(A_{\lambda=2},X_{\lambda=5}\Rightarrow AX=3\mathbf{d}_1 \), \(\{
\(AX=\sqrt{(-3)^2+(-15)^2+(3)^2}\) or \(3\sqrt{27}\ \{=\sqrt{243}\}=9\sqrt{3}\)A1 cao \(9\sqrt{3}\) seen in (c) only
Part (d) Way 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{YA}{``9\sqrt{3}"}=\tan(``74.36964...")\)M1 \(\frac{YA}{\text{their }
\(YA=55.71758...=55.7\ (1\text{ dp})\)A1 anything that rounds to 55.7
Part (e) Way 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\{A_{\lambda=2},X_{\lambda=5}\Rightarrow\) So \(AX=2AB\Rightarrow\) So at \(B\), \(\lambda=3.5\) or \(\lambda=0.5\}\)
\(\overrightarrow{OB}=\begin{pmatrix}4\\28\\4\end{pmatrix}+3.5\begin{pmatrix}-1\\-5\\1\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\)M1 Substitutes either \(\lambda=\frac{(\text{their }\lambda_X\text{ found in }(a))+2}{2}\) or \(\lambda_B=3-\frac{(\text{their }\lambda_X\text{ found in }(a))}{2}\) into \(l_1\)
\(\overrightarrow{OB}=\begin{pmatrix}4\\28\\4\end{pmatrix}+0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\)A1 At least one position vector is correct (also allow coordinates)
Both position vectors correctA1 Also allow coordinates
Part (e) Way 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}+0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\)M1 Applies either \(\overrightarrow{OA}+0.5\overrightarrow{AX}\) or \(\overrightarrow{OA}-0.5\overrightarrow{AX}\) where \((\text{their }\overrightarrow{AX})=\pm[(\text{their }\overrightarrow{OX})-\overrightarrow{OA}]\)
\(\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}-0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\)A1 At least one position vector correct
Both correctA1
Part (e) Way 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{AB}=\begin{pmatrix}4-\lambda\\28-5\lambda\\4+\lambda\end{pmatrix}-\begin{pmatrix}2\\18\\6\end{pmatrix}=\begin{pmatrix}2-\lambda\\10-5\lambda\\-2+\lambda\end{pmatrix}\); \(\overrightarrow{AX}=\begin{pmatrix}-3\\-15\\3\end{pmatrix}\); \(AX^2=243\Rightarrow AB^2=27(2-\lambda)^2\)
\(AX=2AB\Rightarrow AX^2=4AB^2\Rightarrow 243=4(27)(2-\lambda)^2\Rightarrow (2-\lambda)^2=\frac{9}{4}\) or \(27\lambda^2-108\lambda+\frac{189}{4}=0\) or equivalent; \(\Rightarrow\lambda=3.5\) or \(\lambda=0.5\)M1 Full method of solving for \(\lambda\) the equation \(AX^2=4AB^2\) using their \(\overrightarrow{AX}\) and \(\overrightarrow{AB}\) and substitutes at least one value of \(\lambda\) into \(l_1\)
At least one position vector correctA1
Both position vectors correctA1
Part (e) Way 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{OB}=\begin{pmatrix}-1\\3\\9\end{pmatrix}+0.5\begin{pmatrix}3\\15\\-3\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\)M1 Applies either \((\text{their }\overrightarrow{OX})+0.5\overrightarrow{XA}\) or \((\text{their }\overrightarrow{OX})+1.5\overrightarrow{XA}\) where \((\text{their }\overrightarrow{XA})=\overrightarrow{OA}-(\text{their }\overrightarrow{OX})\)
\(\overrightarrow{OB}=\begin{pmatrix}-1\\3\\9\end{pmatrix}+1.5\begin{pmatrix}3\\15\\-3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\)A1 At least one position vector correct
Both correctA1
Part (e) Way 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\overrightarrow{OB}=0.5\left(\begin{pmatrix}-1\\3\\9\end{pmatrix}+\begin{pmatrix}2\\18\\6\end{pmatrix}\right)=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\)M1 Applies \(\frac{1}{2}\left[(\text{their }\overrightarrow{OX})+\overrightarrow{OA}\right]\)
\(\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}-0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\)A1 At least one position vector correct
Both correctA1
Question 6:
Part (e) – Way 6
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\{\overrightarrow{AX}}\ = 9\sqrt{3}\), \(\
\(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} + 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix} \cdot 3 = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}\)M1 Applies either \(\overrightarrow{OA} + 0.5(K\mathbf{d_1})\) or \(\overrightarrow{OA} - 0.5(K\mathbf{d_1})\), where \(K = \frac{\text{their } \
At least one position vector correctA1 Also allow coordinates
\(\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} - 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix} \cdot 3 = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}\)A1 Both position vectors correct (also allow coordinates)
[3]
Part (a) Notes
AnswerMarks
NoteGuidance
M1 can be implied by at least two correct follow through coordinates from their \(\lambda\) or from their \(\mu\)
Part (b) Notes
AnswerMarks
NoteGuidance
*Evaluating* the dot product (i.e. \((-1)(3)+(-5)(0)+(1)(-4)\)) is not required for M1, dM1 marks
For M1 dM1: Allow one slip in writing down direction vectors \(\mathbf{d_1}\) and \(\mathbf{d_2}\)
Allow M1 dM1 for \(\left(\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}\right)\cos\theta = \pm\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}\)
\(\theta = 1.297995...^c\) (without evidence of awrt 74.37) is A0
Part (b) Way 2 – Vector Cross Product
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{d_1} \times \mathbf{d_2} = \begin{pmatrix}-1\\-5\\1\end{pmatrix}\times\begin{pmatrix}3\\0\\-4\end{pmatrix} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&-5&1\\3&0&-4\end{vmatrix} = 20\mathbf{i}-\mathbf{j}+15\mathbf{k}\)M1 Realisation that vector cross product is required between \(\mathbf{d_1}\) and \(\mathbf{d_2}\), or a multiple
\(\sin\theta = \frac{\sqrt{(20)^2+(-1)^2+(15)^2}}{\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}}\)dM1 Applies vector product formula between \(\mathbf{d_1}\) and \(\mathbf{d_2}\) or a multiple
\(\sin\theta = \frac{\sqrt{626}}{\sqrt{27}\cdot\sqrt{25}} \Rightarrow \theta = 74.36964117... = 74.37\) (2 dp)A1 awrt 74.37 seen in (b) only
[3]
Part (c) Notes
AnswerMarks Guidance
NoteGuidance
M1Finds difference between \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\) and applies Pythagoras to find \(AX\) or \(XA\); OR applies \( (\text{their } \lambda_x \text{ found in (a)}) - 2
For M1: Allow one slip in writing down \(\overrightarrow{OX}\) and \(\overrightarrow{OA}\)
Allow M1A1 for \(\begin{pmatrix}3\\15\\3\end{pmatrix}\) leading to \(AX = \sqrt{(3)^2+(15)^2+(3)^2} = \sqrt{243} = 9\sqrt{3}\)
Part (e) Note
AnswerMarks
NoteGuidance
Imply M1 for no working leading to any two components of one of the \(\overrightarrow{OB}\) which are correct
Part (d) Way 2
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{"9\sqrt{3}"}{YA} = \tan(90 - "74.36964...")\)M1 \(\frac{\text{their}\
\(YA = 55.71758... = 55.7\) (1 dp)A1 Anything that rounds to 55.7
[2]
Part (d) Way 3
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{YA}{\sin("74.36964...")} = \frac{"9\sqrt{3}"}{\sin(90-"74.36964...")}\)M1 \(\frac{YA}{\sin\theta} = \frac{\text{their}\
\(YA = \frac{9\sqrt{3}\sin(74.36964...)}{\sin(15.63036...)} = 55.71758... = 55.7\) (1 dp)A1 Anything that rounds to 55.7
[2]
Part (d) Way 4
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{d_1} = \begin{pmatrix}-1\\-5\\1\end{pmatrix}\), \(\overrightarrow{OY} = \begin{pmatrix}5\\3\\1\end{pmatrix}+\mu\begin{pmatrix}3\\0\\-4\end{pmatrix} = \begin{pmatrix}5+3\mu\\3\\1-4\mu\end{pmatrix}\)— —
\(\overrightarrow{YA} = \begin{pmatrix}2\\18\\6\end{pmatrix}-\begin{pmatrix}5+3\mu\\3\\1-4\mu\end{pmatrix} = \begin{pmatrix}-3-3\mu\\15\\5+4\mu\end{pmatrix}\)— —
\(\overrightarrow{YA}\cdot\mathbf{d_1} = 0 \Rightarrow \begin{pmatrix}-3-3\mu\\15\\5+4\mu\end{pmatrix}\cdot\begin{pmatrix}-1\\-5\\1\end{pmatrix} = 0\)M1 Allow sign slip in copying \(\mathbf{d_1}\); Applies \(\overrightarrow{YA}\cdot\mathbf{d_1}=0\) or \(\overrightarrow{AY}\cdot\mathbf{d_1}=0\) or \(\overrightarrow{YA}\cdot(K\mathbf{d_1})=0\) or \(\overrightarrow{AY}\cdot(K\mathbf{d_1})=0\) to find \(\mu\) and applies Pythagoras for \(AY^2\) or distance \(AY\)
\(\Rightarrow 3+3\mu-75+5+4\mu = 0 \Rightarrow \mu = \frac{67}{7}\)— —
\(YA^2 = \left(-3-3\left(\frac{67}{7}\right)\right)^2+(15)^2+\left(5+4\left(\frac{67}{7}\right)\right)^2\)— —
\(YA = \sqrt{\left(-\frac{222}{7}\right)^2+(15)^2+\left(\frac{303}{7}\right)^2} = 55.71758... = 55.7\) (1 dp)A1 Anything that rounds to 55.7
Note: \(\overrightarrow{OY} = \frac{236}{7}\mathbf{i}+3\mathbf{j}-\frac{261}{7}\mathbf{k}\), \(\overrightarrow{AY} = -\frac{222}{7}\mathbf{i}+15\mathbf{j}+\frac{303}{7}\mathbf{k}\)
[2] 
# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{l_1=l_2\Rightarrow\}\ 28-5\lambda=3\ \{\Rightarrow\lambda=5\}$ or $4-\lambda=5+3\mu$ and $4+\lambda=1-4\mu\ \{\Rightarrow\mu=-2\}$ | B1 | $28-5\lambda=3$ or $4-\lambda=5+3\mu$ and $4+\lambda=1-4\mu$; or $\lambda=5$ or $\mu=-2$ (can be implied) |
| $\left\{\overrightarrow{OX}=\right\}\begin{pmatrix}4\\28\\4\end{pmatrix}+5\begin{pmatrix}-1\\-5\\1\end{pmatrix}$ or $\begin{pmatrix}5\\3\\1\end{pmatrix}-2\begin{pmatrix}3\\0\\-4\end{pmatrix}$ | M1 | Puts $l_1=l_2$ and solves to find $\lambda$ and/or $\mu$; **and** substitutes their value for $\lambda$ into $l_1$ or their value for $\mu$ into $l_2$ |
| So, $X(-1,3,9)$ | A1 cao | $(-1,3,9)$ or $\begin{pmatrix}-1\\3\\9\end{pmatrix}$ or $-\mathbf{i}+3\mathbf{j}+9\mathbf{k}$ or condone $\begin{pmatrix}-1\\3\\9\end{pmatrix}$ |

## Part (b) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d}_1=\begin{pmatrix}-1\\-5\\1\end{pmatrix}$, $\mathbf{d}_2=\begin{pmatrix}3\\0\\-4\end{pmatrix}\Rightarrow\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}$ | M1 | Realisation that the dot product is required between $\mathbf{d}_1$ and $\mathbf{d}_2$, or a multiple of $\mathbf{d}_1$ and $\mathbf{d}_2$ |
| $\cos\theta=\frac{\pm\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}}{\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}}\left\{=\frac{-7}{\sqrt{27}\cdot\sqrt{25}}\right\}$ | dM1 | Dependent on 1st M mark; applies dot product formula between $\mathbf{d}_1$ and $\mathbf{d}_2$ or a multiple |
| $\{\theta=105.6303588...\Rightarrow\}\ \theta_{\text{Acute}}=74.36964117...=74.37\ (2\text{ dp})$ | A1 | awrt 74.37 seen in (b) only |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AX}=``\overrightarrow{OX}"-\overrightarrow{OA}=\begin{pmatrix}-1\\3\\9\end{pmatrix}-\begin{pmatrix}2\\18\\6\end{pmatrix}=\begin{pmatrix}-3\\-15\\3\end{pmatrix}$ or $A_{\lambda=2},X_{\lambda=5}\Rightarrow AX=3|\mathbf{d}_1|$, $\{|\mathbf{d}_1|=\sqrt{27}\}$ | M1 | Full method for finding $AX$ or $XA$ |
| $AX=\sqrt{(-3)^2+(-15)^2+(3)^2}$ or $3\sqrt{27}\ \{=\sqrt{243}\}=9\sqrt{3}$ | A1 cao | $9\sqrt{3}$ seen in (c) only |

## Part (d) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{YA}{``9\sqrt{3}"}=\tan(``74.36964...")$ | M1 | $\frac{YA}{\text{their }|\overrightarrow{AX}|}=\tan\theta$ or $YA=(\text{their }|\overrightarrow{AX}|)\tan\theta$, where $\theta$ is their acute or obtuse angle between $l_1$ and $l_2$ |
| $YA=55.71758...=55.7\ (1\text{ dp})$ | A1 | anything that rounds to 55.7 |

## Part (e) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{A_{\lambda=2},X_{\lambda=5}\Rightarrow$ So $AX=2AB\Rightarrow$ So at $B$, $\lambda=3.5$ or $\lambda=0.5\}$ | | |
| $\overrightarrow{OB}=\begin{pmatrix}4\\28\\4\end{pmatrix}+3.5\begin{pmatrix}-1\\-5\\1\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}$ | M1 | Substitutes **either** $\lambda=\frac{(\text{their }\lambda_X\text{ found in }(a))+2}{2}$ **or** $\lambda_B=3-\frac{(\text{their }\lambda_X\text{ found in }(a))}{2}$ into $l_1$ |
| $\overrightarrow{OB}=\begin{pmatrix}4\\28\\4\end{pmatrix}+0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}$ | A1 | At least one position vector is correct (also allow coordinates) |
| Both position vectors correct | A1 | Also allow coordinates |

## Part (e) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}+0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}$ | M1 | Applies either $\overrightarrow{OA}+0.5\overrightarrow{AX}$ **or** $\overrightarrow{OA}-0.5\overrightarrow{AX}$ where $(\text{their }\overrightarrow{AX})=\pm[(\text{their }\overrightarrow{OX})-\overrightarrow{OA}]$ |
| $\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}-0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}$ | A1 | At least one position vector correct |
| Both correct | A1 | |

## Part (e) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB}=\begin{pmatrix}4-\lambda\\28-5\lambda\\4+\lambda\end{pmatrix}-\begin{pmatrix}2\\18\\6\end{pmatrix}=\begin{pmatrix}2-\lambda\\10-5\lambda\\-2+\lambda\end{pmatrix}$; $\overrightarrow{AX}=\begin{pmatrix}-3\\-15\\3\end{pmatrix}$; $AX^2=243\Rightarrow AB^2=27(2-\lambda)^2$ | | |
| $AX=2AB\Rightarrow AX^2=4AB^2\Rightarrow 243=4(27)(2-\lambda)^2\Rightarrow (2-\lambda)^2=\frac{9}{4}$ or $27\lambda^2-108\lambda+\frac{189}{4}=0$ or equivalent; $\Rightarrow\lambda=3.5$ or $\lambda=0.5$ | M1 | Full method of solving for $\lambda$ the equation $AX^2=4AB^2$ using their $\overrightarrow{AX}$ and $\overrightarrow{AB}$ and substitutes at least one value of $\lambda$ into $l_1$ |
| At least one position vector correct | A1 | |
| Both position vectors correct | A1 | |

## Part (e) Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OB}=\begin{pmatrix}-1\\3\\9\end{pmatrix}+0.5\begin{pmatrix}3\\15\\-3\end{pmatrix}=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}$ | M1 | Applies **either** $(\text{their }\overrightarrow{OX})+0.5\overrightarrow{XA}$ **or** $(\text{their }\overrightarrow{OX})+1.5\overrightarrow{XA}$ where $(\text{their }\overrightarrow{XA})=\overrightarrow{OA}-(\text{their }\overrightarrow{OX})$ |
| $\overrightarrow{OB}=\begin{pmatrix}-1\\3\\9\end{pmatrix}+1.5\begin{pmatrix}3\\15\\-3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}$ | A1 | At least one position vector correct |
| Both correct | A1 | |

## Part (e) Way 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OB}=0.5\left(\begin{pmatrix}-1\\3\\9\end{pmatrix}+\begin{pmatrix}2\\18\\6\end{pmatrix}\right)=\begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}$ | M1 | Applies $\frac{1}{2}\left[(\text{their }\overrightarrow{OX})+\overrightarrow{OA}\right]$ |
| $\overrightarrow{OB}=\begin{pmatrix}2\\18\\6\end{pmatrix}-0.5\begin{pmatrix}-3\\-15\\3\end{pmatrix}=\begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}$ | A1 | At least one position vector correct |
| Both correct | A1 | |

# Question 6:

## Part (e) – Way 6

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|{\overrightarrow{AX}}\| = 9\sqrt{3}$, $\|d_1\| = 3\sqrt{3} \Rightarrow K = \frac{9\sqrt{3}}{3\sqrt{3}} = 3 \Rightarrow \overrightarrow{AX} = 3\mathbf{d_1}$ | — | So $\overrightarrow{OB} = \overrightarrow{OA} \pm \frac{1}{2}\overrightarrow{AX} = \overrightarrow{OA} \pm \frac{1}{2}(3\mathbf{d_1})$ |
| $\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} + 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix} \cdot 3 = \begin{pmatrix}0.5\\10.5\\7.5\end{pmatrix}$ | M1 | Applies either $\overrightarrow{OA} + 0.5(K\mathbf{d_1})$ or $\overrightarrow{OA} - 0.5(K\mathbf{d_1})$, where $K = \frac{\text{their } \|\overrightarrow{AX}\|}{3\sqrt{3}}$ |
| At least one position vector correct | A1 | Also allow coordinates |
| $\overrightarrow{OB} = \begin{pmatrix}2\\18\\6\end{pmatrix} - 0.5\begin{pmatrix}-1\\-5\\1\end{pmatrix} \cdot 3 = \begin{pmatrix}3.5\\25.5\\4.5\end{pmatrix}$ | A1 | Both position vectors correct (also allow coordinates) |
| **[3]** | | |

---

## Part (a) Notes

| Note | Guidance |
|---|---|
| M1 can be implied by at least two correct follow through coordinates from their $\lambda$ or from their $\mu$ | |

## Part (b) Notes

| Note | Guidance |
|---|---|
| *Evaluating* the dot product (i.e. $(-1)(3)+(-5)(0)+(1)(-4)$) is not required for M1, dM1 marks | |
| **For M1 dM1:** Allow one slip in writing down direction vectors $\mathbf{d_1}$ and $\mathbf{d_2}$ | |
| Allow M1 dM1 for $\left(\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}\right)\cos\theta = \pm\begin{pmatrix}-1\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-4\end{pmatrix}$ | |
| $\theta = 1.297995...^c$ (without evidence of awrt 74.37) is A0 | |

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## Part (b) Way 2 – Vector Cross Product

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d_1} \times \mathbf{d_2} = \begin{pmatrix}-1\\-5\\1\end{pmatrix}\times\begin{pmatrix}3\\0\\-4\end{pmatrix} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&-5&1\\3&0&-4\end{vmatrix} = 20\mathbf{i}-\mathbf{j}+15\mathbf{k}$ | M1 | Realisation that vector cross product is required between $\mathbf{d_1}$ and $\mathbf{d_2}$, or a multiple |
| $\sin\theta = \frac{\sqrt{(20)^2+(-1)^2+(15)^2}}{\sqrt{(-1)^2+(-5)^2+(1)^2}\cdot\sqrt{(3)^2+(0)^2+(-4)^2}}$ | dM1 | Applies vector product formula between $\mathbf{d_1}$ and $\mathbf{d_2}$ or a multiple |
| $\sin\theta = \frac{\sqrt{626}}{\sqrt{27}\cdot\sqrt{25}} \Rightarrow \theta = 74.36964117... = 74.37$ (2 dp) | A1 | awrt 74.37 seen in (b) only |
| **[3]** | | |

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## Part (c) Notes

| Note | Guidance |
|---|---|
| M1 | Finds difference between $\overrightarrow{OX}$ and $\overrightarrow{OA}$ and applies Pythagoras to find $AX$ or $XA$; **OR** applies $|(\text{their } \lambda_x \text{ found in (a)}) - 2|\cdot\sqrt{(-1)^2+(-5)^2+(1)^2}$ |
| For M1: Allow one slip in writing down $\overrightarrow{OX}$ and $\overrightarrow{OA}$ | |
| Allow M1A1 for $\begin{pmatrix}3\\15\\3\end{pmatrix}$ leading to $AX = \sqrt{(3)^2+(15)^2+(3)^2} = \sqrt{243} = 9\sqrt{3}$ | |

## Part (e) Note

| Note | Guidance |
|---|---|
| Imply M1 for no working leading to any two components of one of the $\overrightarrow{OB}$ which are correct | |

---

## Part (d) Way 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{"9\sqrt{3}"}{YA} = \tan(90 - "74.36964...")$ | M1 | $\frac{\text{their}\|\overrightarrow{AX}\|}{YA} = \tan(90-\theta)$ or $AY = \frac{\text{their}\|\overrightarrow{AX}\|}{\tan(90-\theta)}$, where $\theta$ is acute or obtuse angle between $l_1$ and $l_2$ |
| $YA = 55.71758... = 55.7$ (1 dp) | A1 | Anything that rounds to 55.7 |
| **[2]** | | |

---

## Part (d) Way 3

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{YA}{\sin("74.36964...")} = \frac{"9\sqrt{3}"}{\sin(90-"74.36964...")}$ | M1 | $\frac{YA}{\sin\theta} = \frac{\text{their}\|\overrightarrow{AX}\|}{\sin(90-\theta)}$, where $\theta$ is acute or obtuse angle between $l_1$ and $l_2$ |
| $YA = \frac{9\sqrt{3}\sin(74.36964...)}{\sin(15.63036...)} = 55.71758... = 55.7$ (1 dp) | A1 | Anything that rounds to 55.7 |
| **[2]** | | |

---

## Part (d) Way 4

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d_1} = \begin{pmatrix}-1\\-5\\1\end{pmatrix}$, $\overrightarrow{OY} = \begin{pmatrix}5\\3\\1\end{pmatrix}+\mu\begin{pmatrix}3\\0\\-4\end{pmatrix} = \begin{pmatrix}5+3\mu\\3\\1-4\mu\end{pmatrix}$ | — | — |
| $\overrightarrow{YA} = \begin{pmatrix}2\\18\\6\end{pmatrix}-\begin{pmatrix}5+3\mu\\3\\1-4\mu\end{pmatrix} = \begin{pmatrix}-3-3\mu\\15\\5+4\mu\end{pmatrix}$ | — | — |
| $\overrightarrow{YA}\cdot\mathbf{d_1} = 0 \Rightarrow \begin{pmatrix}-3-3\mu\\15\\5+4\mu\end{pmatrix}\cdot\begin{pmatrix}-1\\-5\\1\end{pmatrix} = 0$ | M1 | Allow sign slip in copying $\mathbf{d_1}$; Applies $\overrightarrow{YA}\cdot\mathbf{d_1}=0$ or $\overrightarrow{AY}\cdot\mathbf{d_1}=0$ or $\overrightarrow{YA}\cdot(K\mathbf{d_1})=0$ or $\overrightarrow{AY}\cdot(K\mathbf{d_1})=0$ to find $\mu$ and applies Pythagoras for $AY^2$ or distance $AY$ |
| $\Rightarrow 3+3\mu-75+5+4\mu = 0 \Rightarrow \mu = \frac{67}{7}$ | — | — |
| $YA^2 = \left(-3-3\left(\frac{67}{7}\right)\right)^2+(15)^2+\left(5+4\left(\frac{67}{7}\right)\right)^2$ | — | — |
| $YA = \sqrt{\left(-\frac{222}{7}\right)^2+(15)^2+\left(\frac{303}{7}\right)^2} = 55.71758... = 55.7$ (1 dp) | A1 | Anything that rounds to 55.7 |
| **Note:** $\overrightarrow{OY} = \frac{236}{7}\mathbf{i}+3\mathbf{j}-\frac{261}{7}\mathbf{k}$, $\overrightarrow{AY} = -\frac{222}{7}\mathbf{i}+15\mathbf{j}+\frac{303}{7}\mathbf{k}$ | | |
| **[2]** | | |

---
6. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations

$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 
4 \\
28 \\
4
\end{array} \right) + \lambda \left( \begin{array} { r } 
- 1 \\
- 5 \\
1
\end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 
5 \\
3 \\
1
\end{array} \right) + \mu \left( \begin{array} { r } 
3 \\
0 \\
- 4
\end{array} \right)$$

where $\lambda$ and $\mu$ are scalar parameters.

The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the point $X$.
\item Find the size of the acute angle between $l _ { 1 }$ and $l _ { 2 }$, giving your answer in degrees to 2 decimal places.

The point $A$ lies on $l _ { 1 }$ and has position vector $\left( \begin{array} { r } 2 \\ 18 \\ 6 \end{array} \right)$
\item Find the distance $A X$, giving your answer as a surd in its simplest form.

The point $Y$ lies on $l _ { 2 }$. Given that the vector $\overrightarrow { Y A }$ is perpendicular to the line $l _ { 1 }$
\item find the distance $Y A$, giving your answer to one decimal place.

The point $B$ lies on $l _ { 1 }$ where $| \overrightarrow { A X } | = 2 | \overrightarrow { A B } |$.
\item Find the two possible position vectors of $B$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4 2017 Q6 [13]}}
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Q1 8 Q2 6 Q3 12 Q4 9 Q5 7 Q6 13 Q7 8 Q8 12