Home / IBDP Maths AA: Topic : AHL 3.15: Coincident, parallel, intersecting and skew lines: IB style Questions HL Paper 2

# IBDP Maths AA: Topic : AHL 3.15: Coincident, parallel, intersecting and skew lines: IB style Questions HL Paper 2

## 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]

## 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]

a.

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 – 2y – 3z = 2 and 2xyz = k

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

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

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

x = t,= (2 − 3k) + 5t and z = (2− 2) − 3t     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]

b.

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]

c.

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]

d.

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, M1 for attempt to use perpendicular condition to determine relation between parameters, M1 for attempt to use the area to find the parameters and A2 for final answer.

[5 marks]

e.

## 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.

a.

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.

b.

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.

c.

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.

d.

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.

e.

## 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]

## 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]

a.

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, A1 for the first factor and A1 for the complete second factor.

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, A1 for $${\left( {12 – \frac{{5t}}{3}} \right)^2}$$ and A1 for $${(5 – 4t)^2}$$.

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 L1 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 L2 by $${{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}} 3\\ { – 3}\\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 1}\\ 3\\ { – 4} \end{array}} \right)$$ .

The lines L1 and L2 intersect at point T. Find the coordinates of T.

## 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.

## 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.

## 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]

## 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]

a.

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]

b.

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]

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