Question
Consider the planes \({\pi _1}:x – 2y – 3z = 2{\text{ and }}{\pi _2}:2x – y – z = k\) .
Find the angle between the planes \({\pi _1}\)and \({\pi _2}\) .
The planes \({\pi _1}\) and \({\pi _2}\) intersect in the line \({L_1}\) . Show that the vector equation of
\({L_1}\) is \(r = \left( {\begin{array}{*{20}{c}}
0\\
{2 – 3k}\\
{2k – 2}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
5\\
{ – 3}
\end{array}} \right)\)
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.
Determine a Cartesian equation of the plane \({\pi _3}\) containing both lines \({L_1}\) and \({L_2}\) .
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.
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 – 2y – 3z = 2 and 2x – y – z = k
\(5x – y = 3k – 2 \Rightarrow x = \frac{{y – (2 – 3k)}}{5}\) M1A1
eliminate y from x – 2y – 3z = 2 and 2x – y – z = k
\(3x + z = 2k – 2 \Rightarrow x = \frac{{z – (2k – 2)}}{{ – 3}}\) A1
x = t, y = (2 − 3k) + 5t and z = (2k − 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]
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, 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]
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.
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.
Question
Consider the points P(−3, −1, 2) and Q(5, 5, 6).
Find a vector equation for the line, \({L_1}\), which passes through the points P and Q.
The line \({L_2}\) has equation
\[r = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
0 \\
4
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
5 \\
2 \\
0
\end{array}} \right){\text{.}}\]
Show that \({L_1}\) and \({L_2}\) intersect at the point R(1, 2, 4).
Find the acute angle between \({L_1}\) and \({L_2}\).
Let S be a point on \({L_2}\) such that \(\left| {\overrightarrow {{\text{RP}}} } \right| = \left| {\overrightarrow {{\text{RS}}} } \right|\).
Show that one of the possible positions for S is \({{\text{S}}_1}\)(−4, 0, 4) and find the coordinates of the other possible position, \({{\text{S}}_2}\).
Let S be a point on \({L_2}\) such that \(\left| {\overrightarrow {{\text{RP}}} } \right| = \left| {\overrightarrow {{\text{RS}}} } \right|\).
Find a vector equation of the line which passes through R and bisects \({\rm{P\hat R}}{{\text{S}}_1}\).
Answer/Explanation
Markscheme
\(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}}
8 \\
6 \\
4
\end{array}} \right)\) (A1)
equation of line: \(r = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
{ – 1} \\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
8 \\
6 \\
4
\end{array}} \right)\) (or equivalent) M1A1
Note: Award M1A0 if r = is omitted.
[3 marks]
METHOD 1
\(x:{\text{ }}- 4 + 5s = – 3 + 8t\)
\(y:{\text{ }}2s = – 1 + 6t\)
\(z:{\text{ }}4 = 2 + 4t\) M1
solving any two simultaneously M1
t = 0.5, s = 1 (or equivalent) A1
verification that these values give R when substituted into both equations (or that the three equations are consistent and that one gives R) R1
METHOD 2
(1, 2, 4) is given by t = 0.5 for \({L_1}\) and s = 1 for \({L_2}\) M1A1A1
because (1, 2, 4) is on both lines it is the point of intersection of the two lines R1
[4 marks]
\(\left( {\begin{array}{*{20}{c}}
5 \\
2 \\
0
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
4 \\
3 \\
2
\end{array}} \right) = 26 = \sqrt {29} \times \sqrt {29} \cos \theta \) M1
\(\cos \theta = \frac{{26}}{{29}}\) (A1)
\(\theta = 0.459\) or 26.3° A1
[3 marks]
\(\overrightarrow {{\text{RP}}} = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
{ – 1} \\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 3} \\
{ – 2}
\end{array}} \right)\), \(\left| {\overrightarrow {{\text{RP}}} } \right| = \sqrt {29} \) (M1)A1
Note: This could also be obtained from \(\left| {0.5\left( {\begin{array}{*{20}{c}}
8 \\
6 \\
4
\end{array}} \right)} \right|\)
EITHER
\(\overrightarrow {{\text{R}}{{\text{S}}_{\text{1}}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
0 \\
4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 5} \\
{ – 2} \\
0
\end{array}} \right)\), \(\left| {\overrightarrow {{\text{R}}{{\text{S}}_{\text{1}}}} } \right| = \sqrt {29} \) A1
\(\therefore \overrightarrow {{\text{O}}{{\text{S}}_2}} = \overrightarrow {{\text{O}}{{\text{S}}_{\text{1}}}} + 2\overrightarrow {{{\text{S}}_{\text{1}}}{\text{R}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
0 \\
4
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
5 \\
2 \\
0
\end{array}} \right)\) M1A1
\(\left( {{\text{or }}\overrightarrow {{\text{O}}{{\text{S}}_2}} = \overrightarrow {{\text{OR}}} + \overrightarrow {{{\text{S}}_{\text{1}}}{\text{R}}} = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5 \\
2 \\
0
\end{array}} \right)} \right)\)
\( = \left( {\begin{array}{*{20}{c}}
6 \\
4 \\
4
\end{array}} \right)\)
\({{\text{S}}_2}\) is (6, 4, 4) A1
OR
\(\left( {\begin{array}{*{20}{c}}
{ – 4 + 5s} \\
{2s} \\
4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{5s – 5} \\
{2s – 2} \\
0
\end{array}} \right)\) M1
\({(5s – 5)^2} + {(2s – 2)^2} = 29\) M1A1
\(29{s^2} – 58s + 29 = 29\)
\(s(s – 2) = 0,{\text{ }}s = 0,{\text{ 2}}\)
\((6,{\text{ }}4,{\text{ 4}}){\text{ }}\left( {{\text{and (}} – 4,{\text{ }}0,{\text{ }}4{\text{)}}} \right)\) A1
Note: There are several geometrical arguments possible using information obtained in previous parts, depending on what forms the previous answers had been given.
[6 marks]
EITHER
midpoint of \([{\text{P}}{{\text{S}}_1}]\) is M(–3.5, –0.5, 3) M1A1
\(\overrightarrow {{\text{RM}}} = \left( {\begin{array}{*{20}{c}}
{ – 4.5} \\
{ – 2.5} \\
{ – 1}
\end{array}} \right)\) A1
OR
\(\overrightarrow {{\text{R}}{{\text{S}}_1}} = \left( {\begin{array}{*{20}{c}}
{ – 5} \\
{ – 2} \\
0
\end{array}} \right)\) M1
the direction of the line is \({\overrightarrow {{\text{RS}}} _1} + \overrightarrow {{\text{RP}}} \)
\(\left( {\begin{array}{*{20}{c}}
{ – 5} \\
{ – 2} \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 3} \\
{ – 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 9} \\
{ – 5} \\
{ – 2}
\end{array}} \right)\) M1A1
THEN
the equation of the line is:
\(r = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
9 \\
5 \\
2
\end{array}} \right)\) or equivalent A1
Note: Marks cannot be awarded for methods involving halving the angle, unless it is clear that the candidate considers also the equation of the plane of \({L_1}\) and \({L_2}\) to reduce the number of parameters involved to one (to obtain the vector equation of the required line).
[4 marks]
Examiners report
There were many good answers to part (a) showing a clear understanding of finding the vector equation of a line. Unfortunately this understanding was marred by many students failing to write the equation properly resulting in just 2 marks out of the 3. The most common response was of the form \({L_1} = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
{ – 1} \\
2
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
4 \\
3 \\
2
\end{array}} \right)\) which seemed a waste of a mark.
In part (b) many students failed to verify that the lines do indeed intersect.
Part (c) was very well done.
In part (d) most candidates were able to obtain the first three marks, but few were able to find the second point.
There were few correct answers to part (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, \({{r}} = {{a}} + t{{b}}\),
find the vectors \({{a}}\) and \({{b}}\).
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.
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, 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]
Examiners report
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