IB DP Maths Topic 4.4 Finding the point of intersection of two lines. SL Paper 2

 

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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.

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 \({L_1}\) passes through points \({\text{A}}(1{\text{, }} – 1{\text{, }}4)\) and \({\text{B}}(2{\text{, }} – 2{\text{, }}5)\) .

Line \({L_2}\) has equation \({\boldsymbol{r}} = \left( \begin{array}{l}
2\\
4\\
7
\end{array} \right) + s\left( \begin{array}{l}
2\\
1\\
3
\end{array} \right)\) .

Find \(\overrightarrow {{\rm{AB}}} \) .

[2]
a.

Find an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .

[2]
b.

Find the angle between \({L_1}\) and \({L_2}\) .

[7]
c.

The lines \({L_1}\) and \({L_2}\) intersect at point C. Find the coordinates of C.

[6]
d.
Answer/Explanation

Markscheme

appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(B – A\)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\)     A1     N2

[2 marks]

a.

any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)     A2 N2

where \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{2 + t}\\
{ – 2 – t}\\
{5 + t}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} – 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} – {\boldsymbol{j}} + {\boldsymbol{k}})\)

 [2 marks]

b.

choosing correct direction vectors \(\left( \begin{array}{l}
2\\
1\\
3
\end{array} \right)\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\)     (A1)(A1)

finding scalar product and magnitudes     (A1)(A1)(A1)

scalar product \( = 1 \times 2 + – 1 \times 1 + 1 \times 3\) \(\left( { = 4} \right)\)

magnitudes \(\sqrt {{1^2} + {{( – 1)}^2} + {1^2}}{\text{  }}( = 1.73 \ldots )\) , \(\sqrt {4 + 1 + 9}{\text{  }}( = 3.74 \ldots )\)

substitution into \(\frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) (accept \(\theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) , but not \(\sin \theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) )     M1

e.g. \(\cos \theta = \frac{{1 \times 2 + – 1 \times 1 + 1 \times 3}}{{\sqrt {{1^2} + {{( – 1)}^2} + {1^2}} \sqrt {{2^2} + {1^2} + {3^2}} }}\) , \(\cos \theta = \frac{4}{{\sqrt {42} }}\)

\(\theta = 0.906\) \(({51.9^ \circ })\)     A1      N5

[7 marks]

c.

METHOD 1 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)t = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations    A1A1

e.g. \(1 + t = 2 + 2s\) , \( – 1 – t = 4 + s\) , \(4 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = – 3\) , \(s = – 2\)

C is \(( – 2{\text{, }}2{\text{, }}1)\)     A1     N3

METHOD 2 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 2} \\
  5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 2}\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations     A1A1

e.g. \(2 + t = 2 + 2s\) , \( – 2 – t = 4 + s\) , \(5 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = – 4\) , \(s = – 2\)

C is \(( – 2{\text{, }}2{\text{, }}1)\)     A1     N3

[6 marks]

d.

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 \(\rm{P}\).

Find the coordinates of \({\text{P}}\).

[6]
a.

Show that the lines are perpendicular.

[5]
b.

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]
c.
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]

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]

c.
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