Download IITian Academy App for accessing Online Mock Tests and More..
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
The line \({L_1}\) is represented by the vector equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\) .
A second line \({L_2}\) is parallel to \({L_1}\) and passes through the point B(\( – 8\), \( – 5\), \(25\)) .
Write down a vector equation for \({L_2}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
A third line \({L_3}\) is perpendicular to \({L_1}\) and is represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) .
Show that \(k = – 2\) .
The lines \({L_1}\) and \({L_3}\) intersect at the point A.
Find the coordinates of A.
The lines \({L_2}\)and \({L_3}\)intersect at point C where \(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
6\\
3\\
{ – 24}
\end{array}} \right)\) .
(i) Find \(\overrightarrow {{\rm{AB}}} \) .
(ii) Hence, find \(|\overrightarrow {{\rm{AC}}} |\) .
Answer/Explanation
Markscheme
any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\)
Note: Award A1 for \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[2 marks]
recognizing scalar product must be zero (seen anywhere) R1
e.g. \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\)
evidence of choosing direction vectors \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) (A1)(A1)
correct calculation of scalar product (A1)
e.g. \(2( – 7) + 1( – 2) – 8k\)
simplification that clearly leads to solution A1
e.g. \( – 16 – 8k\) , \( – 16 – 8k = 0\)
\(k = – 2\) AG N0
[5 marks]
evidence of equating vectors (M1)
e.g. \({L_1} = {L_3}\) , \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
{ – 2}
\end{array}} \right)\)
any two correct equations A1A1
e.g. \( – 3 + 2p = 5 – 7q\) , \( – 1 + p = – 2q\) , \(- 25 – 8p = 3 – 2q\)
attempting to solve equations (M1)
finding one correct parameter (\(p = – 3\) , \(q = 2\) ) A1
the coordinates of A are \(( – 9, – 4, – 1)\) A1 N3
[6 marks]
(i) evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{ – 9}\\
{ – 4}\\
{ – 1}
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{26}
\end{array}} \right)\) A1 N2
(ii) finding \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
7\\
2\\
2
\end{array}} \right)\) A1
evidence of finding magnitude (M1)
e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{7^2} + {2^2} + {2^2}} \)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {57} \) A1 N3
[5 marks]
Question
The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .
(i) Show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{BD}}} \) .
(iii) Show that \(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) .
The line (AC) has equation \({\boldsymbol{r}} = {\boldsymbol{u}} + s{\boldsymbol{v}}\) .
(i) Write down vector u and vector v .
(ii) Find a vector equation for the line (BD).
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Show that \(k = 1\) .
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Hence find the area of triangle ACD.
Answer/Explanation
Markscheme
(i) correct approach A1
e.g. \(\overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OA}}} \) , \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\)
\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) AG N0
(ii) appropriate approach (M1)
e.g. \({\mathop{\rm D}\nolimits} – {\rm{B}}\) , \(\left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right)\) , move 3 to the right and 6 down
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) A1 N2
(iii) finding the scalar product A1
e.g. \(4(3) + 2( – 6)\) , \(12 – 12\)
valid reasoning R1
e.g. \(4(3) + 2( – 6) = 0\) , scalar product is zero
\(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) AG N0
[5 marks]
(i) correct “position” vector for u; “direction” vector for v A1A1 N2
e.g. \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\) , \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\) ; \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) , \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 1}
\end{array}} \right)\)
accept in equation e.g. \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 4}\\
{ – 2}
\end{array}} \right)\)
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , where \({\boldsymbol{b}} = \overrightarrow {{\rm{BD}}} \)
\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 1}\\
2
\end{array}} \right)\) A2 N2
[4 marks]
METHOD 1
substitute (3, k) into equation for (AC) or (BD) (M1)
e.g. \(3 = 1 + 4s\) , \(3 = 1 + 3t\)
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \(k = 0 + \frac{1}{2}(2)\)
\(k = 1\) AG N0
METHOD 2
setting up two equations (M1)
e.g. \(1 + 4s = 4 + 3t\) , \(2s = – 1 – 6t\) ; setting vector equations of lines equal
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \frac{1}{3}\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\)
\(k = 1\) AG N0
[3 marks]
\(\overrightarrow {{\rm{PD}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}
\end{array}} \right)\) (A1)
\(|\overrightarrow {{\rm{PD}}} | = \sqrt {{2^2} + {1^2}} \) \(( = \sqrt 5 )\) (A1)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {{4^2} + {2^2}} \) \(( = \sqrt {20} )\) (A1)
area \( = \frac{1}{2} \times |\overrightarrow {{\rm{AC}}} | \times |\overrightarrow {{\rm{PD}}} |\) (\( = \frac{1}{2} \times \sqrt {20} \times \sqrt 5 \)) M1
\( = 5\) A1 N4
[5 marks]
| Marks available | 4 |
| Reference code | 11N.1.sl.TZ0.8 |
Question
The line \({L_1}\) passes through the points P(2, 4, 8) and Q(4, 5, 4) .
The line \({L_2}\) is perpendicular to \({L_1}\) , and parallel to \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\) , where \(p \in \mathbb{Z}\) .
(i) Find \(\overrightarrow {{\rm{PQ}}} \) .
(ii) Hence write down a vector equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) .
(i) Find the value of p .
(ii) Given that \({L_2}\) passes through \({\text{R}}(10{\text{, }}6{\text{, }}- 40)\) , write down a vector equation for \({L_2}\) .
The lines \({L_1}\) and \({L_2}\) intersect at the point A. Find the x-coordinate of A.
Answer/Explanation
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{P}} – {\rm{Q}}\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A1 N2
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) (accept any parameter for s)
where a is \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}
4\\
5\\
4
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4 + 2s}\\
{5 + 1s}\\
{4 – 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} + 4{\boldsymbol{j}} + 8{\boldsymbol{k}} + s(2{\boldsymbol{i}} + 1{\boldsymbol{j}} – 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + s{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + s{\boldsymbol{a}}\) .
[4 marks]
(i) choosing correct direction vectors for \({L_1}\) and \({L_2}\) (A1) (A1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\)
evidence of equating scalar product to 0 (M1)
correct calculation of scalar product A1
e.g. \(2 \times 3p + 1 \times 2p + ( – 4) \times 4\) , \(8p – 16 = 0\)
\(p = 2\) A1 N3
(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for t)
where a is \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10 + 6s}\\
{6 + 4s}\\
{ – 40 + 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 10{\boldsymbol{i}} + 6{\boldsymbol{j}} – 40{\boldsymbol{k}} + s(6{\boldsymbol{i}} + 4{\boldsymbol{j}} + 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (unless they have been penalised for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) in part (a)), A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[7 marks]
appropriate approach (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\)
any two correct equations with different parameters A1A1
e.g. \(2 + 2s = 10 + 6t\) , \(4 + s = 6 + 4t\) , \(8 – 4s = – 40 + 4t\)
attempt to solve simultaneous equations (M1)
correct working (A1)
e.g. \( – 6 = – 2 – 2t\) , \(4 = 2t\) , \( – 4 + 5s = 46\) , \(5s = 50\)
one correct parameter \(s = 10\) , \(t = 2\) A1
\(x = 22\) (accept (22, 14, −32)) A1 N4
[7 marks]
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 passes through the point \((5, – 4,10)\) and is parallel to the vector \(\left( {\begin{array}{*{20}{c}}
4\\
{ – 2}\\
5
\end{array}} \right)\) .
Write down a vector equation for line L .
The line L intersects the x-axis at the point P. Find the x-coordinate of P.
Answer/Explanation
Markscheme
any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for t)
where a is \(\left( {\begin{array}{*{20}{c}}
5\\
{ – 4}\\
{10}
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
4\\
{ – 2}\\
5
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
5 \\
{ – 4} \\
{10}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
4 \\
{ – 2} \\
5
\end{array}} \right){\text{, }}{\boldsymbol{r}} = 5{\boldsymbol{i}} – 4{\boldsymbol{j}} + 10{\boldsymbol{k}} + t( – 8{\boldsymbol{i}} + 4{\boldsymbol{j}} – 10{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[2 marks]
recognizing that \(y = 0\) or \(z = 0\) at x-intercept (seen anywhere) (R1)
attempt to set up equation for x-intercept (must suggest \(x \ne 0\) ) (M1)
e.g. \(L = \left( {\begin{array}{*{20}{c}}
x\\
0\\
0
\end{array}} \right)\) , \(5 + 4t = x\) , \(r = \left( {\begin{array}{*{20}{c}}
1\\
0\\
0
\end{array}} \right)\)
one correct equation in one variable (A1)
e.g. \( – 4 – 2t = 0\) , \(10 + 5t = 0\)
finding \(t = – 2\) A1
correct working (A1)
e.g. \(x = 5 + ( – 2)(4)\)
\(x = – 3\) (accept \(( – 3{\text{, }}0{\text{, }}0)\)) A1 N3
[6 marks]
Question
Consider points A(\(1\), \( – 2\), \( -1\)) , B(\(7\), \( – 4\), \(3\)) and C(\(1\), \( -2\), \(3\)) . The line \({L_1}\) passes through C and is parallel to \(\overrightarrow {{\rm{AB}}} \) .
A second line, \({L_2}\) , is given by \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
p
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{AB}}} \) .
Hence, write down a vector equation for \({L_1}\) .
Given that \({L_1}\) is perpendicular to \({L_2}\) , show that \(p = – 6\) .
The line \({L_1}\) intersects the line \({L_2}\) at point Q. Find the \(x\)-coordinate of Q.
Answer/Explanation
Markscheme
valid approach (M1)
eg \(\left( {\begin{array}{*{20}{c}}
7\\
{ – 4}\\
3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) , \({\rm{A}} – {\rm{B}}\) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) A1 N2
[2 marks]
any correct equation in the form \(\boldsymbol{r} = \boldsymbol{a} + t\boldsymbol{b}\) (accept any parameter for \(t\))
where \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) and \(\boldsymbol{b}\) is a scalar multiple of \(\overrightarrow {{\rm{AB}}} \) A2 N2
eg \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \((x,y,z) = (1, – 2,3) + t(3, – 1,2)\) , \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{1 + 6t}\\
{ – 2 – 2t}\\
{3 + 4t}
\end{array}} \right)\)
Note: Award A1 for \(\boldsymbol{a} + t\boldsymbol{b}\) , A1 for \({L_1} = \boldsymbol{a} + t\boldsymbol{b}\) , A0 for \(\boldsymbol{r} = \boldsymbol{b} + t\boldsymbol{a}\) .
[2 marks]
recognizing that scalar product \( = 0\) (seen anywhere) R1
correct calculation of scalar product (A1)
eg \(6(3) – 2( – 3) + 4p\) , \(18 + 6 + 4p\)
correct working A1
eg \(24 + 4p = 0\) , \(4p = – 24\)
\(p = – 6\) AG N0
[3 marks]
setting lines equal (M1)
eg \({L_1} = {L_2}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
{ – 6}
\end{array}} \right)\)
any two correct equations with different parameters A1A1
eg \(1 + 6t = 1 + 3s\) , \( – 2 – 2t = 2 – 3s\) , \(3 + 4t = 15 – 6s\)
attempt to solve their simultaneous equations (M1)
one correct parameter A1
eg \(t = \frac{1}{2}\) , \(s = \frac{5}{3}\)
attempt to substitute parameter into vector equation (M1)
eg \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \(1 + \frac{1}{2} \times 6\)
\(x = 4\) (accept (4, -3, 5), ignore incorrect values for \(y\) and \(z\)) 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]
Question
Distances in this question are in metres.
Ryan and Jack have model airplanes, which take off from level ground. Jack’s airplane takes off after Ryan’s.
The position of Ryan’s airplane \(t\) seconds after it takes off is given by \(\boldsymbol{r}=\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\).
Find the speed of Ryan’s airplane.
Find the height of Ryan’s airplane after two seconds.
The position of Jack’s airplane \(s\) seconds after it takes off is given by r = \(\left( \begin{array}{c} – 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\).
Show that the paths of the airplanes are perpendicular.
The two airplanes collide at the point \((-23, 20, 28)\).
How long after Ryan’s airplane takes off does Jack’s airplane take off?
Answer/Explanation
Markscheme
valid approach (M1)
eg magnitude of direction vector
correct working (A1)
eg \(\sqrt {{{( – 4)}^2} + {2^2} + {4^2}} ,{\text{ }}\sqrt { – {4^2} + {2^2} + {4^2}} \)
\(6{\text{ (m}}{{\text{s}}^{ – 1}})\) A1 N2
[3 marks]
substituting \(2\) for \(t\) (A1)
eg \(0 + 2(4)\), r = \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + 2\left( \begin{array}{c} – 4\\2\\4\end{array} \right),\left( \begin{array}{c} – 3\\10\\8\end{array} \right)\), \(y = 10\)
\(8\) (metres) A1 N2
[2 marks]
METHOD 1
choosing correct direction vectors \(\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\) (A1)(A1)
evidence of scalar product M1
eg a \( \cdot \) b
correct substitution into scalar product (A1)
eg \(( – 4 \times 4) + (2 \times – 6) + (4 \times 7)\)
evidence of correct calculation of the scalar product as \(0\) A1
eg \( – 16 – 12 + 28 = 0\)
directions are perpendicular AG N0
METHOD 2
choosing correct direction vectors \(\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\) (A1)(A1)
attempt to find angle between vectors M1
correct substitution into numerator A1
eg \(\cos \theta = \frac{{ – 16 – 12 + 28}}{{\left| a \right|\left| b \right|}},{\text{ }}\cos \theta = 0\)
\(\theta = 90^\circ \) A1
directions are perpendicular AG N0
[5 marks]
METHOD 1
one correct equation for Ryan’s airplane (A1)
eg \(5 – 4t = – 23,{\text{ }}6 + 2t = 20,{\text{ }}0 + 4t = 28\)
\(t = 7\) A1
one correct equation for Jack’s airplane (A1)
eg \( – 39 + 4s = – 23,{\text{ }}44 – 6s = 20,{\text{ }}0 + 7s = 28\)
\(s = 4\) A1
\(3\) (seconds later) A1 N2
METHOD 2
valid approach (M1)
eg \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} – 4\\2\\4\end{array} \right) = \left( \begin{array}{c} – 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\), one correct equation
two correct equations (A1)
eg \(5 – 4t = – 39 + 4s,{\text{ }}6 + 2t = 44 – 6s,{\text{ }}4t = 7s\)
\(t = 7\) A1
\(s = 4\) A1
\(3\) (seconds later) A1 N2
[5 marks]
Question
A line \({L_1}\) passes through the points \({\text{A}}(0,{\text{ }} – 3,{\text{ }}1)\) and \({\text{B}}( – 2,{\text{ }}5,{\text{ }}3)\).
(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\).
(ii) Write down a vector equation for \({L_1}\).
A line \({L_2}\) has equation \({\mathbf{r}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 7 \\ { – 4} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { – 1} \end{array}} \right)\). The lines \({L_1}\) and \({L_2}\) intersect at a point \(C\).
Show that the coordinates of \(C\) are \(( – 1,{\text{ }}1,{\text{ }}2)\).
A point \(D\) lies on line \({L_2}\) so that \(\left| {\overrightarrow {{\text{CD}}} } \right| = \sqrt {18} \) and \(\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} = – 9\). Find \({\rm{A\hat CD}}\).
Answer/Explanation
Markscheme
(i) correct approach A1
eg\(\;\;\;{\text{OB}} – {\text{OA, }}\left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right),{\text{ B}} – {\text{A}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\) AG N0
(ii) any correct equation in the form \(r = a + \) t\(b\) (accept any parameter for \(t\))
where \(a\) is \(\left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right)\), and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\) A2 N2
eg\(r\) = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),\(r\) = \left( {\begin{array}{*{20}{c}} { – 2 – 2s} \\ {5 + 8s} \\ {3 + 2s} \end{array}} \right),\(r = 2i + 5j + 3k + \) t\(( – 2i + 8j + 2k)\)
Note: Award A1 for the form \(a\) + t\(b\), A1 for the form \(L = \(a\) + t\(b\),
A0 for the form \(r\) = \(b\) + t\(a\).
[3 marks]
valid approach (M1)
eg\(\;\;\;\)equating lines, \({L_1} = {L_2}\)
one correct equation in one variable A1
eg\(\;\;\; – 2t = – 1,{\text{ }} – 2 – 2t = – 1\)
valid attempt to solve (M1)
eg\(\;\;\;2t = 1,{\text{ }} – 2t = 1\)
one correct parameter A1
eg\(\;\;\;t = \frac{1}{2},{\text{ }}t = – \frac{1}{2},{\text{ }}s = – 6\)
correct substitution of either parameter A1
eg\(\;\;\;r = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right) – \frac{1}{2}\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { – 1} \\ 7 \\ { – 4} \end{array}} \right) – 6\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { – 1} \end{array}} \right)\)
the coordinates of \(C\) are \(( – 1,{\text{ }}1,{\text{ }}2)\), or position vector of \(C\) is \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 2 \end{array}} \right)\) AG N0
Note: If candidate uses the same parameter in both vector equations and working shown, award M1A1M1A0A0.
[5 marks]
valid approach (M1)
eg\(\;\;\;\)attempt to find \(\overrightarrow {{\text{CA}}} ,{\text{ }}\cos {\rm{A\hat CD}} = \frac{{\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} }}{{\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CD}}} } \right|}},{\rm{ A\hat CD}}\) formed by \(\overrightarrow {{\text{CA}}} \) and \(\overrightarrow {{\text{CD}}} \)
\(\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 4} \\ { – 1} \end{array}} \right)\) (A1)
Notes: Exceptions to FT:
1 if candidate indicates that they are finding \(\overrightarrow {{\text{CA}}} \), but makes an error, award M1A0;
2 if candidate finds an incorrect vector (including \(\overrightarrow {{\text{AC}}} \)), award M0A0.
In both cases, if working shown, full FT may be awarded for subsequent correct FT work.
Award the final (A1) for simplification of their value for \({\rm{A\hat CD}}\).
Award the final A2 for finding their arc cos. If their value of cos does not allow them to find an angle, they cannot be awarded this A2.
finding \(\left| {\overrightarrow {{\text{CA}}} } \right|\) (may be seen in cosine formula) A1
eg\(\;\;\;\sqrt {{1^2} + {{( – 4)}^2} + {{( – 1)}^2}} ,{\text{ }}\sqrt {18} \)
correct substitution into cosine formula (A1)
eg\(\;\;\;\frac{{ – 9}}{{\sqrt {18} \sqrt {18} }}\)
finding \(\cos {\rm{A\hat CD}} – \frac{1}{2}\) (A1)
\({\rm{A\hat CD}} = \frac{{2\pi }}{3}\;\;\;(120^\circ )\) A2 N2
Notes: Award A1 if additional answers are given.
Award A1 for answer \(\frac{\pi }{3}{\rm{ (60^\circ )}}\).
[7 marks]
Total [15 marks]