## Question

Consider the planes \({\pi _1}:x – 2y – 3z = 2{\text{ and }}{\pi _2}:2x – y – z = k\) .

a.Find the angle between the planes \({\pi _1}\)and \({\pi _2}\) .[4]

b.The planes \({\pi _1}\) and \({\pi _2}\) intersect in the line \({L_1}\) . Show that the vector equation of

\({L_1}\) is \(r = \left( {\begin{array}{*{20}{c}}

0\\

{2 – 3k}\\

{2k – 2}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\)[5]

c.The line \({L_2}\) has Cartesian equation \(5 – x = y + 3 = 2 – 2z\) . The lines \({L_1}\) and \({L_2}\) intersect at a point X. Find the coordinates of X.[5]

d.Determine a Cartesian equation of the plane \({\pi _3}\) containing both lines \({L_1}\) and \({L_2}\) .[5]

e.Let Y be a point on \({L_1}\) and Z be a point on \({L_2}\) such that XY is perpendicular to YZ and the area of the triangle XYZ is 3. Find the perimeter of the triangle XYZ.[5]

**▶️Answer/Explanation**

## Markscheme

**Note: **Accept alternative notation for vectors (*eg* \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).

\(\boldsymbol{n} = \left( {\begin{array}{*{20}{c}}

1 \\

{ – 2} \\

{ – 3}

\end{array}} \right)\) and \(\boldsymbol{m} = \left( {\begin{array}{*{20}{c}}

2 \\

{ – 1} \\

{ – 1}

\end{array}} \right)\) *(A1)*

\(\cos \theta = \frac{{\boldsymbol{n} \cdot \boldsymbol{m}}}{{\left| \boldsymbol{n} \right|\left| \boldsymbol{m} \right|}}\) *(M1)*

\(\cos \theta = \frac{{2 + 2 + 3}}{{\sqrt {1 + 4 + 9} \sqrt {4 + 1 + 1} }} = \frac{7}{{\sqrt {14} \sqrt 6 }}\) *A1*

\(\theta = 40.2^\circ \,\,\,\,\,(0.702{\text{ rad}})\) *A1*

*[4 marks]*

**Note: **Accept alternative notation for vectors (*eg* \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).

**METHOD 1**

eliminate *z *from *x* – 2*y* – 3*z* = 2 and 2*x* – *y* – *z* = *k*

\(5x – y = 3k – 2 \Rightarrow x = \frac{{y – (2 – 3k)}}{5}\) *M1A1*

eliminate *y *from *x* – 2*y* – 3*z* = 2 and 2*x* – *y* – *z* = *k*

\(3x + z = 2k – 2 \Rightarrow x = \frac{{z – (2k – 2)}}{{ – 3}}\) *A1*

*x* =* t*,* y *= (2 − 3*k*) + 5t and *z* = (2*k *− 2) − 3*t A1A1*

\(r = \left( {\begin{array}{*{20}{c}}

0\\

{2 – 3k}\\

{2k – 2}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\) *AG*

*[5 marks]*

**METHOD 2**

\(\left( {\begin{array}{*{20}{c}}

1\\

{ – 2}\\

{ – 3}

\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}

2\\

{ – 1}\\

{ – 1}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

{ – 1}\\

{ – 5}\\

3

\end{array}} \right) \Rightarrow {\text{direction is }}\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\) ** M1A1**

Let *x* = 0

\(0 – 2y – 3z = 2{\text{ and }}2 \times 0 – y – z = k\) **(M1)**

solve simultaneously **(M1)**

\(y = 2 – 3k{\text{ and }}z = 2k – 2\) **A1**

therefore **r** \( = \left( {\begin{array}{*{20}{c}}

0\\

{2 – 3k}\\

{2k – 2}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\) **AG**

**[5 marks]**

**METHOD 3**

substitute \(x = t,{\text{ }}y = (2 – 3k) + 5t{\text{ and }}z = (2k – 2) – 3t{\text{ into }}{\pi _1}{\text{ and }}{\pi _2}\) ** M1**

for \({\pi _1}:t – 2(2 – 3k + 5t) – 3(2k – 2 – 3t) = 2\) **A1**

for \({\pi _2}:2t – (2 – 3k + 5t) – (2k – 2 – 3t) = k\) **A1**

the planes have a unique line of intersection **R2**

therefore the line is \(r = \left( {\begin{array}{*{20}{c}}

0\\

{2 – 3k}\\

{2k – 2}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\) **AG**

**[5 marks]**

**Note: **Accept alternative notation for vectors (*eg* \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).

\(5 – t = (2 – 3k + 5t) + 3 = 2 – 2(2k – 2 – 3t)\) **M1A1**

**Note: **Award ** M1A1 **if candidates use vector or parametric equations of \({L_2}\)

*eg *\(\left( {\begin{array}{*{20}{c}}

0\\

{2 – 3k}\\

{2k – 2}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

5\\

{ – 3}\\

1

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

{ – 2}\\

2\\

{ – 1}

\end{array}} \right)\) or \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}

{t = 5 – 2s}\\

{2 – 3k + 5t = – 3 + 2s}\\

{2k – 2 – 3t = 1 + s}

\end{array}} \right.\)

solve simultaneously *M1*

\(k = 2,{\text{ }}t = 1{\text{ }}(s = 2)\) **A1**

intersection point (\(1\), \(1\), \( – 1\)) *A1*

* *

*[5 marks]*

**Note: **Accept alternative notation for vectors (*eg* \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).

\({\overrightarrow l _2} = \left( {\begin{array}{*{20}{c}}

2\\

{ – 2}\\

1

\end{array}} \right)\) **A1**

\({\overrightarrow l _1} \times {\overrightarrow l _2} = \left| {\begin{array}{*{20}{c}}

\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\

1&5&{ – 3}\\

2&{ – 2}&1

\end{array}} \right| = \left( {\begin{array}{*{20}{c}}

{ – 1}\\

{ – 7}\\

{ – 12}

\end{array}} \right)\) **(M1)A1**

\(\boldsymbol{r} \cdot \left( {\begin{array}{*{20}{c}}

1\\

7\\

{12}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

1\\

1\\

{ – 1}

\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}

1\\

7\\

{12}

\end{array}} \right)\) **(M1)**

\(x + 7y + 12z = – 4\) **A1**

**[5 marks]**

**Note: **Accept alternative notation for vectors (*eg* \(\langle a{\text{, }}b{\text{, }}c\rangle {\text{ or }}\left( {a{\text{, }}b{\text{, }}c} \right)\)).

Let \(\theta \) be the angle between the lines \({\overrightarrow l _1} = \left( {\begin{array}{*{20}{c}}

1\\

5\\

{ – 3}

\end{array}} \right)\) and \({\overrightarrow l _2} = \left( {\begin{array}{*{20}{c}}

2\\

{ – 2}\\

1

\end{array}} \right)\)

\(\cos \theta = \frac{{\left| {2 – 10 – 3} \right|}}{{\sqrt {35} \sqrt 9 }} \Rightarrow \theta = 0.902334…{\text{ }}51.699…^\circ )\) *(M1)*

as the triangle XYZ has a right angle at Y,

\({\text{XZ}} = a \Rightarrow {\text{YZ}} = a\sin \theta {\text{ and XY}} = a\cos \theta \) *(M1)*

\({\text{area = 3}} \Rightarrow \frac{{{a^2}\sin \theta \cos \theta }}{2} = 3\) *(M1)*

\(a = 3.5122…\) **(A1)**

perimeter \( = a + a\sin \theta + a\cos \theta = 8.44537… = 8.45\) **A1**

**Note: **If candidates attempt to find coordinates of Y and Z award ** M1 **for expression of vector YZ in terms of two parameters,

**for attempt to use perpendicular condition to determine relation between parameters,**

*M1***for attempt to use the area to find the parameters and**

*M1***for final answer.**

*A2*

*[5 marks]*

## Examiners report

Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of *k*. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.

Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.

Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of *k*. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.

Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.

Although this was the last question in part B, it was answered surprisingly well by many candidates, except for part (e). Even those who had not done so well elsewhere often gained a number of marks in some parts of the question. Nevertheless the presence of parameters seemed to have blocked the abilities of weaker candidates to solve situations in which vectors were involved. Mathematical skills for this particular question were sometimes remarkable, however, calculations proved incomplete due to the way that planes were presented. Most candidates found a correct angle in part (a). Occasional arithmetic errors in calculating the magnitude of a vector and dot product occurred. In part (b) the vector product approach was popular. In some case candidates simply verified the result by substitution. There was a lot of simultaneous equation solving, much of which was not very pretty. In part (c), a number of candidates made errors when attempting to solve a system of equations involving parameters. Many of the results for the point were found in terms of *k*. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.

Very few excellent answers to part (e) were seen using an efficient method. Most candidates attempted methods involving heavy algebraic manipulation and had little success in this part of the question.

*k*. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.

*k*. It was notorious that candidates did not use their GDC to try to find the coordinates of the intersection point between lines. In part (d), a number of candidates used an incorrect point but this part was often done well.

## Question

Ed walks in a straight line from point \({\text{P}}( – 1,{\text{ }}4)\) to point \({\text{Q}}(4,{\text{ }}16)\) with constant speed.

Ed starts from point \(P\) at time \(t = 0\) and arrives at point \(Q\) at time \(t = 3\), where \(t\) is measured in hours.

Given that, at time \(t\), Ed’s position vector, relative to the origin, can be given in the form, \({{r}} = {{a}} + t{{b}}\),

a.find the vectors \({{a}}\) and \({{b}}\).[3]

b.Roderick is at a point \({\text{C}}(11,{\text{ }}9)\). During Ed’s walk from \(P\) to \(Q\) Roderick wishes to signal to Ed. He decides to signal when Ed is at the closest point to \(C\).

Find the time when Roderick signals to Ed.[5]

**▶️Answer/Explanation**

## Markscheme

\({{a}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)\) *A1*

\({{b}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 4 \\ {16} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} {\frac{5}{3}} \\ 4 \end{array}} \right)\) *(M1)A1*

*[3 marks]*

**METHOD 1**

Roderick must signal in a direction vector perpendicular to Ed’s path. *(M1)*

the equation of the signal is \({\mathbf{s}} = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\;\;\;\)(or equivalent) *A1*

\(\left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right) + \frac{t}{3}\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\) *M1*

\(\frac{5}{3}t + 12\lambda = 12\) and \(4t – 5\lambda = 5\) *M1*

\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) *A1*

**METHOD 2**

\(\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) \bullet \left( {\left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1 + \frac{5}{3}t} \\ {4 + 4t} \end{array}} \right)} \right) = 0\;\;\;\)(or equivalent) *M1A1A1*

**Note: **Award the ** M1 **for an attempt at a scalar product equated to zero,

**for the first factor and**

*A1***for the complete second factor.**

*A1*attempting to solve for \(t\) *(M1)*

\(t = 2.13\;\;\;\left( {\frac{{360}}{{169}}} \right)\) *A1*

**METHOD 3**

\(x = \sqrt {{{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \;\;\;\)(or equivalent)\(\;\;\;\left( {{x^2} = {{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \right)\) *M1A1A1*

**Note: **Award ** M1 **for use of Pythagoras’ theorem,

**for \({\left( {12 – \frac{{5t}}{3}} \right)^2}\) and**

*A1***for \({(5 – 4t)^2}\).**

*A1*attempting (graphically or analytically) to find \(t\) such that \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0\left( {\frac{{{\text{d}}({x^2})}}{{{\text{d}}t}} = 0} \right)\) *(M1)*

\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) *A1*

**METHOD 4**

\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)}}{{\left| {\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right)} \right|\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|}} = \frac{{120}}{{169}}\) *M1A1*

**Note: **Award ** M1 **for attempting to calculate the scalar product.

\(\frac{{120}}{{13}} = \frac{t}{3}\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|\;\;\;\)(or equivalent) *(A1)*

attempting to solve for \(t\) *(M1)*

\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) *A1*

*[5 marks]*

*Total [8 marks]*

## Question

The line *L _{1}* is represented by \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}

2\\

5\\

3

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

1\\

2\\

3

\end{array}} \right)\) and the line

*L*by \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}

_{2}3\\

{ – 3}\\

8

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

{ – 1}\\

3\\

{ – 4}

\end{array}} \right)\) .

The lines *L _{1}* and

*L*intersect at point T. Find the coordinates of T.

_{2}**▶️Answer/Explanation**

## Markscheme

evidence of equating vectors **(M1)**

e.g. \({L_1} = {L_2}\)

for any **two** correct equations **A1A1**

e.g. \(2 + s = 3 – t\) , \(5 + 2s = – 3 + 3t\) , \(3 + 3s = 8 – 4t\)

attempting to solve the equations **(M1)**

finding **one** correct parameter \((2 = – 1{\text{, }}t = 2)\) **A1**

the coordinates of T are \((1{\text{, }}3{\text{, }}0)\) **A1 N3**

** [6 marks]**

## Question

Two lines with equations \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}

2\\

3\\

{ – 1}

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

5\\

{ – 3}\\

2

\end{array}} \right)\) and \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}

9\\

2\\

2

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

{ – 3}\\

5\\

{ – 1}

\end{array}} \right)\) intersect at the point P. Find the coordinates of P.

**▶️Answer/Explanation**

## Markscheme

evidence of appropriate approach **(M1)**

e.g. \(\left( {\begin{array}{*{20}{c}}

2\\

3\\

{ – 1}

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

5\\

{ – 3}\\

2

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

9\\

2\\

2

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

{ – 3}\\

5\\

{ – 1}

\end{array}} \right)\)

two correct equations **A1A1 **

e.g. \(2 + 5s = 9 – 3t\) , \(3 – 3s = 2 + 5t\) , \( – 1 + 2s = 2 – t\)

attempting to solve the equations **(M1)**

one correct parameter \(s = 2\) , \(t = – 1\) **A1**

P is \((12, – 3,3)\) (accept \(\left( {\begin{array}{*{20}{c}}

{12}\\

{ – 3}\\

3

\end{array}} \right)\)) **A1 N3**

* [6 marks] *

## Question

Line \({L_1}\) has equation \({\boldsymbol{r}_1} = \left( {\begin{array}{*{20}{c}}

{10}\\

6\\

{ – 1}

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

2\\

{ – 5}\\

{ – 2}

\end{array}} \right)\) and line \({L_2}\) has equation \({\boldsymbol{r}_2} = \left( {\begin{array}{*{20}{c}}

2\\

1\\

{ – 3}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

3\\

5\\

2

\end{array}} \right)\) .

Lines \({L_1}\) and \({L_2}\) intersect at point A. Find the coordinates of A.

**▶️Answer/Explanation**

## Markscheme

appropriate approach ** (M1)**

*eg * \(\left( {\begin{array}{*{20}{c}}

{10}\\

6\\

{ – 1}

\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}

2\\

{ – 5}\\

{ – 2}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

2\\

1\\

{ – 3}

\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}

3\\

5\\

2

\end{array}} \right)\) , \({L_1} = {L_2}\)

any two correct equations ** A1A1**

*eg * \(10 + 2s = 2 + 3t\) , \(6 – 5s = 1 + 5t\) , \( – 1 – 2s = – 3 + 2t\)

attempt to solve **(M1)**

*eg *substituting one equation into another

one correct parameter **A1**

*eg * \(s = – 1\) , \(t = 2\)

correct substitution ** (A1)**

*eg * \(2 + 3(2)\) , \(1 + 5(2)\) , \( – 3 + 2(2)\)

A \( = \) (\(8\), \(11\), \(1\)) (accept column vector) **A1 N4 **

**[7 marks]**

## Question

Consider the lines \({L_1}\) and \({L_2}\) with equations \({L_1}\) : \(\boldsymbol{r}=\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right)\) and \({L_2}\) : \(\boldsymbol{r} = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\).

The lines intersect at point \(\rm{P}\).

a.Find the coordinates of \({\text{P}}\).[6]

b.Show that the lines are perpendicular.[5]

c.The point \({\text{Q}}(7, 5, 3)\) lies on \({L_1}\). The point \({\text{R}}\) is the reflection of \({\text{Q}}\) in the line \({L_2}\).

Find the coordinates of \({\text{R}}\).[6]

**▶️Answer/Explanation**

## Markscheme

appropriate approach *(M1)*

*eg* \(\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\), \({L_1} = {L_2}\)

any **two **correct equations *A1A1*

*eg* \(11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 – s = – 7 + 11t\)

attempt to solve system of equations *(M1)*

*eg *\(10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s – 2t = – 10} \\ {3s – t = – 7} \end{array}} \right.\)

one correct parameter *A1*

*eg* \(s = – 2,{\text{ }}t = 1\)

\({\text{P}}(3, 2, 4)\) (accept position vector) *A1 N3*

*[6 marks]*

choosing correct direction vectors for \({L_1}\) and \({L_2}\) *(A1)(A1)*

*eg* \(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ {11} \end{array}} \right)\) (or any scalar multiple)

evidence of scalar product (with any vectors) *(M1)*

*eg* \(a \cdot b\), \(\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)\)

correct substitution *A1*

*eg* \(4(2) + 3(1) + ( – 1)(11),{\text{ }}8 + 3 – 11\)

calculating \(a \cdot b = 0\) *A1*

**Note: **Do not award the final ** A1 **without evidence of calculation.

vectors are perpendicular *AG N0*

*[5 marks]*

**Note: **Candidates may take different approaches, which do not necessarily involve vectors.

In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.

**METHOD 1**

attempt to find \(\overrightarrow {{\text{QP}}} \) or \(\overrightarrow {{\text{PQ}}} \) *(M1)*

correct working (may be seen on diagram) *A1*

*eg* \(\overrightarrow {{\text{QP}}} \) = \(\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right)\), \(\overrightarrow {{\text{PQ}}} \) = \(\left( \begin{array}{c}7\\5\\3\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right)\)

recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) *(R1)*

*eg* on diagram

\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) **(R1)**

*eg* \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{PR}}} \), marked on diagram

correct working *(A1)*

*eg* \(\left( \begin{array}{c}3\\2\\4\end{array} \right) – \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)\)

\({\text{R}}(–1, –1, 5)\) (accept position vector) *A1 N3*

**METHOD 2 **

recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) *(R1)*

*eg* on diagram

\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) *(R1)*

*eg* \({\text{P}}\) midpoint of \({\text{QR}}\), marked on diagram

valid approach to find **one **coordinate of mid-point *(M1)*

*eg* \({x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)\)

**one **correct substitution *A1*

*eg* \({x_R} = 3 + (3 – 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)\)

correct working for one coordinate *(A1)*

*eg* \({x_R} = 3 – 4,{\text{ }}4 – 5 = {y_R},{\text{ }}8 = (z + 3)\)

\({\text{R}} (-1, -1, 5)\) (accept position vector) *A1 N3*

*[6 marks]*