Question
A line \({L_1}\) passes though points P(−1, 6, −1) and Q(0, 4, 1) .
A second line \({L_2}\) has equation \(r = \left( {\begin{array}{*{20}{c}}
4\\
2\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
0\\
{ – 4}
\end{array}} \right)\) .
(i) Show that \(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\) .
(ii) Hence, write down an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
Find the cosine of the angle between \(\overrightarrow {{\rm{PQ}}} \) and \({L_2}\) .
The lines \({L_1}\) and \({L_2}\) intersect at the point R. Find the coordinates of R.
Answer/Explanation
Markscheme
(i) evidence of correct approach A1
e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} – \overrightarrow {{\rm{OP}}} \) , \(Q – P\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\) AG N0
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) A2 N2
where a is either \(\overrightarrow {{\rm{OP}}} \) or \(\overrightarrow {{\rm{OQ}}} \) and b is a scalar multiple of \(\overrightarrow {{\rm{PQ}}} \)
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
6\\
{ – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
t \\
{4 – 2t} \\
{1 + 2t}
\end{array}} \right)\), \({\boldsymbol{r}} = 4{\boldsymbol{j}} + {\boldsymbol{k}} + t({\boldsymbol{i}} – 2{\boldsymbol{j}} + 2{\boldsymbol{k}})\)
[3 marks]
choosing a correct direction vector for \({L_2}\) (A1)
e.g. \(\left( {\begin{array}{*{20}{c}}
3\\
0\\
{ – 4}
\end{array}} \right)\)
finding scalar products and magnitudes (A1)(A1)(A1)
scalar product \( = 1(3) – 2(0) + 2( – 4)\) \(( = – 5)\)
magnitudes \( = \sqrt {{1^2} + {{( – 2)}^2} + {2^2}} \) \(( = 3)\) , \(\sqrt {{3^2} + {0^2} + {{( – 4)}^2}} \) \(( = 5)\)
substitution into formula M1
e.g. \(\cos \theta = \frac{{ – 5}}{{\sqrt 9 \times \sqrt {25} }}\)
\(\cos \theta = – \frac{1}{3}\) A2 N5
[7 marks]
evidence of valid approach (M1)
e.g. equating lines, \({L_1} = {L_2}\)
EITHER
one correct equation in one variable A2
e.g. \(6 – 2t = 2\)
OR
two correct equations in two variables A1A1
e.g. \(2t + 4s = 0\) , \(t – 3s = 5\)
THEN
attempt to solve (M1)
one correct parameter A1
e.g. \(t = 2\) , \(s = – 1\)
correct substitution of either parameter (A1)
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
4\\
2\\
{ – 1}
\end{array}} \right) + ( – 1)\left( {\begin{array}{*{20}{c}}
3\\
0\\
{ – 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
6\\
{ – 1}
\end{array}} \right) + ( + 2)\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\)
coordinates \({\text{R}}(1{\text{, }}2{\text{, }}3)\) A1 N3
[7 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]