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Question
A particle is moving with a constant velocity along line L . Its initial position is A(6 , −2 , 10) . After one second the particle has moved to B( 9, −6 , 15) .
(i) Find the velocity vector, \(\overrightarrow {{\rm{AB}}} \) .
(ii) Find the speed of the particle.
Write down an equation of the line L .
Answer/Explanation
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \({\rm{B}} – {\rm{A}}\) , \(\left( {\begin{array}{*{20}{c}}
{9 – 6}\\
{ – 6 + 2}\\
{15 – 10}
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\) (accept \(\left( {3, – 4, 5} \right)\) ) A1 N2
(ii) evidence of finding the magnitude of the velocity vector M1
e.g. \({\text{speed}} = \sqrt {{3^2} + {4^2} + {5^2}} \)
\({\text{speed}} = \sqrt {50} \) \(\left( { = 5\sqrt 2 } \right)\) A1 N1
[4 marks]
correct equation (accept Cartesian and parametric forms) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
{10}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
9\\
{ – 6}\\
{15}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\)
[2 marks]
Question
The vertices of the triangle PQR are defined by the position vectors
\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 3}\\
1
\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
2
\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 1}\\
5
\end{array}} \right)\) .
Find
(i) \(\overrightarrow {{\rm{PQ}}} \) ;
(ii) \(\overrightarrow {{\rm{PR}}} \) .
Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .
(i) Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .
(ii) Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .
Answer/Explanation
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} – {\rm{P}}\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
1
\end{array}} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{PR}}} = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\) A1 N1
[3 marks]
METHOD 1
choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \) (A1)(A1)
finding \(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\) (A1) (A1)(A1)
\(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} = – 2 + 4 + 4( = 6)\)
\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( – 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)
substituting into formula for angle between two vectors M1
e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6 \times \sqrt {24} }}\)
simplifying to expression clearly leading to \(\frac{1}{2}\) A1
e.g. \(\frac{6}{{\sqrt 6 \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) AG N0
METHOD 2
evidence of choosing cosine rule (seen anywhere) (M1)
\(\overrightarrow {{\rm{QR}}} = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\) A1
\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \) (A1)(A1)(A1)
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} – {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6 \times \sqrt {24} }}\) A1
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 – 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\) A1
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) AG N0
[7 marks]
(i) METHOD 1
evidence of appropriate approach (M1)
e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram
substituting correctly (A1)
e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} \)
\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\) A1 N3
METHOD 2
since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \) (A1)
evidence of approach
e.g. drawing a right triangle, finding the missing side (A1)
\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\) A1 N3
(ii) evidence of appropriate approach (M1)
e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)
correct substitution
e.g. area \( = \frac{1}{2}\sqrt 6 \times \sqrt {24} \times \frac{{\sqrt 3 }}{2}\) A1
area \( = 3\sqrt 3 \) A1 N2
[6 marks]
Question
The line \({L_1}\) is parallel to the z-axis. The point P has position vector \(\left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right)\) and lies on \({L_1}\).
Write down the equation of \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\).
The line \({L_2}\) has equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\) . The point A has position vector \(\left( {\begin{array}{*{20}{c}}
6\\
2\\
9
\end{array}} \right)\) .
Show that A lies on \({L_2}\) .
Let B be the point of intersection of lines \({L_1}\) and \({L_2}\) .
(i) Show that \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}
8\\
1\\
{14}
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{AB}}} \) .
The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.
Find \(\overrightarrow {{\rm{OD}}} \) .
Answer/Explanation
Markscheme
\({L_1}:{\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
0\\
0\\
1
\end{array}} \right)\) A2 N2
[2 marks]
evidence of equating \({\boldsymbol{r}}\) and \(\overrightarrow {{\rm{OA}}} \) (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
6\\
2\\
9
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\) , \(A = r\)
one correct equation A1
e.g. \(6 = 2 + 2s\) , \(2 = 4 – s\) , \(9 = – 1 + 5s\)
\(s = 2\) A1
evidence of confirming for other two equations A1
e.g. \(6 = 2 + 4\) , \(2 = 4 – 2\) , \(9 = – 1 + 10\)
so A lies on \({L_2}\) AG N0
[4 marks]
(i) evidence of approach M1
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
0\\
0\\
1
\end{array}} \right)\) \({L_1} = {L_2}\)
one correct equation A1
e.g. \(2 + 2s = 8\) , \(4 – s = 1\) , \( – 1 + 5s = t\)
attempt to solve (M1)
finding \(s = 3\) A1
substituting M1
e.g. \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\)
\(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}
8\\
1\\
{14}
\end{array}} \right)\) AG N0
(ii) evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} – \overrightarrow {{\rm{OA}}} \)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\) A1 N2
[7 marks]
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{DC}}} \)
correct values A1
e.g. \(\overrightarrow {{\rm{OD}}} + \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{2 – x}\\
{1 – y}\\
{ – 4 – z}
\end{array}} \right)\)
\(\overrightarrow {{\rm{OD}}} = \left( {\begin{array}{*{20}{c}}
0\\
2\\
{ – 9}
\end{array}} \right)\) A1 N2
[3 marks]
Question
Let \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{BC}}} \) .
Find a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .
Show that \(\overrightarrow {{\rm{AB}}} \) is perpendicular to \(\overrightarrow {{\rm{AC}}} \) .
Answer/Explanation
Markscheme
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 1}\\
{ – 1}
\end{array}} \right)\) A1 N2
[2 marks]
attempt to find the length of \(\overrightarrow {{\rm{AB}}} \) (M1)
\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( – 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\) (A1)
unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}
{\frac{6}{7}}\\
{ – \frac{2}{7}}\\
{\frac{3}{7}}
\end{array}} \right)} \right)\) A1 N2
[3 marks]
recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular (M1)
correct substitution in a scalar product formula A1
e.g. \((6) \times ( – 2) + ( – 2) \times ( – 3) + (3) \times (2)\) , \(\cos \theta = \frac{{ – 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)
correct calculation A1
e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)
therefore, they are perpendicular AG N0
[3 marks]
Question
In the following diagram, \(\overrightarrow {{\text{OP}}} \) = p, \(\overrightarrow {{\text{OQ}}} \) = q and \(\overrightarrow {{\text{PT}}} = \frac{1}{2}\overrightarrow {{\text{PQ}}} \).
Express each of the following vectors in terms of p and q,
\(\overrightarrow {{\text{QP}}} \);
\(\overrightarrow {{\text{OT}}} \).
Answer/Explanation
Markscheme
appropriate approach (M1)
eg \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{QO}}} + \overrightarrow {{\text{OP}}} ,{\text{ P}} – {\text{Q}}\)
\(\overrightarrow {{\text{QP}}} \) = p – q A1 N2
[2 marks]
recognizing correct vector for \(\overrightarrow {{\text{QT}}} \) or \(\overrightarrow {{\text{PT}}} \) (A1)
eg \(\overrightarrow {{\text{QT}}} = \frac{1}{2}\)(p – q), \(\overrightarrow {{\text{PT}}} = \frac{1}{2}\)(q – p)
appropriate approach (M1)
eg \(\overrightarrow {{\text{OT}}} = \overrightarrow {{\text{OP}}} + \overrightarrow {{\text{PT}}} ,{\text{ }}\overrightarrow {{\text{OQ}}} + \overrightarrow {{\text{QT}}} ,{\text{ }}\overrightarrow {{\text{OP}}} + \frac{1}{2}\overrightarrow {{\text{PQ}}} \)
\(\overrightarrow {{\text{OT}}} = \frac{1}{2}\)(p + q) \(\left( {{\text{accept }}\frac{{p + q}}{2}} \right)\) A1 N2
[3 marks]
Question
The line \({L_1}\) passes through the points \(\rm{A}(2, 1, 4)\) and \(\rm{B}(1, 1, 5)\).
Another line \({L_2}\) has equation r = \(\left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + s\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\) . The lines \({L_1}\) and \({L_2}\) intersect at the point P.
Show that \(\overrightarrow {{\text{AB}}} = \) \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)
Hence, write down a direction vector for \({L_1}\);
Hence, write down a vector equation for \({L_1}\).
Find the coordinates of P.
Write down a direction vector for \({L_2}\).
Hence, find the angle between \({L_1}\) and \({L_2}\).
Answer/Explanation
Markscheme
correct approach A1
eg \(\left( \begin{array}{c}1\\1\\5\end{array} \right) – \left( \begin{array}{l}2\\1\\4\end{array} \right)\), \({\text{AO}} + {\text{OB}}\), \(b – a\)
\(\overrightarrow {{\text{AB}}} = \) \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\) AG N0
[1 mark]
correct vector (or any multiple) A1 N1
eg d = \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)
[1 mark]
any correct equation in the form r = a + tb (accept any parameter for t)
where a is \(\left( \begin{array}{c}2\\1\\4\end{array} \right)\) or \(\left( \begin{array}{c}1\\1\\5\end{array} \right)\) , and b is a scalar multiple of \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\) A2 N2
eg r = \(\left( \begin{array}{c}1\\1\\5\end{array} \right) + t\left( \begin{array}{c} – 1\\0\\1\end{array} \right),\left( \begin{array}{c}x\\y\\z\end{array} \right) = \left( \begin{array}{c}2 – s\\1\\4 + s\end{array} \right)\)
Note: Award A1 for a + tb, A1 for \({L_1}\) = a + tb, A0 for r = b + ta.
[2 marks]
valid approach (M1)
eg \({r_1} = {r_2}\), \(\left( \begin{array}{c}2\\1\\4\end{array} \right) + t\left( \begin{array}{c} – 1\\0\\1\end{array} \right) = \left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + s\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\)
one correct equation in one parameter A1
eg \(2 – t = 4, 1 = 7 – s, 1 – t = 4\)
attempt to solve (M1)
eg \(2 – 4 = t, s = 7 – 1, t = 1 – 4\)
one correct parameter A1
eg \(t = -2, s = 6, t = -3\),
attempt to substitute their parameter into vector equation (M1)
eg \(\left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + 6\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\)
P(4, 1, 2) (accept position vector) A1 N2
[6 marks]
correct direction vector for \({L_2}\) A1 N1
eg \(\left( \begin{array}{c}0\\-1\\ 1\end{array} \right)\), \(\left( \begin{array}{c}0\\2\\ – 2\end{array} \right)\)
[1 mark]
correct scalar product and magnitudes for their direction vectors (A1)(A1)(A1)
scalar product \( = 0 \times – 1 + – 1 \times 0 + 1 \times 1{\text{ }}( = 1)\)
magnitudes \( = \sqrt {{0^2} + {{( – 1)}^2} + {1^2}} ,{\text{ }}\sqrt { – {1^2} + {0^2} + {1^2}} \left( {\sqrt 2 ,{\text{ }}\sqrt 2 } \right)\)
attempt to substitute their values into formula M1
eg \(\frac{{0 + 0 + 1}}{{\left( {\sqrt {{0^2} + {{( – 1)}^2} + {1^2}} } \right) \times \left( {\sqrt { – {1^2} + {0^2} + {1^2}} } \right)}},{\text{ }}\frac{1}{{\sqrt 2 \times \sqrt 2 }}\)
correct value for cosine, \(\frac{1}{2}\) A1
angle is \(\frac{\pi }{3}{\text{ }}( = {60^ \circ })\) A1 N1
[6 marks]