Question
Consider the points A(1, −1, 4), B (2, − 2, 5) and O(0, 0, 0).
(a) Calculate the cosine of the angle between \(\overrightarrow {{\text{OA}}} \) and \(\overrightarrow {{\text{AB}}} \).
(b) Find a vector equation of the line \({L_1}\) which passes through A and B.
The line \({L_2}\) has equation r = 2i + 4j + 7k + t(2i + j + 3k), where \(t \in \mathbb{R}\) .
(c) Show that the lines \({L_1}\) and \({L_2}\) intersect and find the coordinates of their point of intersection.
(d) Find the Cartesian equation of the plane which contains both the line \({L_2}\) and the point A.
▶️Answer/Explanation
Markscheme
(a) Use of \(\cos \theta = \frac{{\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} }}{{\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{AB}}} } \right|}}\) (M1)
\({\overrightarrow {{\text{AB}}} }\) = i − j + k A1
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 3 \) and \(\left| {\overrightarrow {{\text{OA}}} } \right| = 3\sqrt 2 \) A1
\(\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} = 6\) A1
substituting gives \(\cos \theta = \frac{2}{{\sqrt 6 }}\,\,\,\,\,\left( { = \frac{{\sqrt 6 }}{3}} \right)\,\,\,\,\,\)or equivalent M1 N1
[5 marks]
(b) \({L_1}\) : r = \(\overrightarrow {{\text{OA}}} + s\overrightarrow {{\text{AB}}} \,\,\,\,\,\) or equivalent (M1)
\({L_1}\) : r = i − j + 4k + s(i − j + k)\(\,\,\,\,\,\)or equivalent A1
Note: Award (M1)A0 for omitting “ r = ” in the final answer.
[2 marks]
(c) Equating components and forming equations involving s and t (M1)
1 + s = 2 + 2t , −1 − s = 4 + t , 4 + s = 7 + 3t
Having two of the above three equations A1A1
Attempting to solve for s or t (M1)
Finding either s = −3 or t = −2 A1
Explicitly showing that these values satisfy the third equation R1
Point of intersection is (−2, 2, 1) A1 N1
Note: Position vector is not acceptable for final A1.
[7 marks]
(d) METHOD 1
\(r = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1} \\
4
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
2 \\
1 \\
3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
{ – 3} \\
3 \\
{ – 3}
\end{array}} \right)\) (A1)
\(x = 1 + 2\lambda – 3\mu {\text{ , }}y = – 1 + \lambda + 3\mu {\text{ and }}z = 4 + 3\lambda – 3\mu \) M1A1
Elimination of the parameters M1
\(x + y = 3\lambda {\text{ so }}4(x + y) = 12\lambda {\text{ and }}y + z = 4\lambda + 3{\text{ so }}3(y + z) = 12\lambda + 9\)
\(3(y + z) = 4(x + y) + 9\) A1
Cartesian equation of plane is 4x + y − 3z = −9 (or equivalent) A1 N1
[6 marks]
METHOD 2
EITHER
The point (2, 4, 7) lies on the plane.
The vector joining (2, 4, 7) and (1, −1, 4) and 2i + j + 3k are parallel to the plane. So they are perpendicular to the normal to the plane.
(i − j + 4k) − (2i + 4j + 7k) = −i − 5j − 3k (A1)
\(n = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 5}&{ – 3} \\
2&1&3
\end{array}} \right|\) M1
= −12i − 3j + 9k\(\,\,\,\,\,\)or equivalent parallel vector A1
OR
\({L_1}\) and \({L_2}\) intersect at D(−2, 2, 1)
\(n = \left| {\begin{array}{*{20}{c}}
i&j&k \\
2&1&{ – 3} \\
{ – 3}&3&{ – 3}
\end{array}} \right|\) M1
= −12i − 3j + 9k\(\,\,\,\,\,\)or equivalent parallel vector A1
THEN
r\( \cdot \)n = (i − j + 4k)\( \cdot \)(−12i − 3j + 9k) M1
= 27 A1
Cartesian equation of plane is 4x + y −3z = −9 (or equivalent) A1 N1
[6 marks]
Total [20 marks]
Question
Two boats, A and B , move so that at time t hours, their position vectors, in kilometres, are r\(_A\) = (9t)i + (3 – 6t)j and r\(_B\) = (7 – 4t)i + (7t – 6)j .
a.Find the coordinates of the common point of the paths of the two boats.[4]
b.Show that the boats do not collide.[2]
▶️Answer/Explanation
Markscheme
METHOD 1
\(9{t_A} = 7 – 4{t_B}\) and
\(3 – 6{t_A} = – 6 + 7{t_B}\) M1A1
solve simultaneously
\({t_A} = \frac{1}{3},{\text{ }}{t_B} = 1\) A1
Note: Only need to see one time for the A1.
therefore meet at (3, 1) A1
[4 marks]
METHOD 2
path of A is a straight line: \(y = – \frac{2}{3}x + 3\) M1A1
Note: Award M1 for an attempt at simultaneous equations.
path of B is a straight line: \(y = – \frac{7}{4}x + \frac{{25}}{4}\) A1
\( – \frac{2}{3}x + 3 = – \frac{7}{4}x + \frac{{25}}{4}{\text{ }}( \Rightarrow x = 3)\)
so the common point is (3, 1) A1
[4 marks]
METHOD 1
boats do not collide because the two times \(\left( {{t_A} = \frac{1}{3},{\text{ }}{t_B} = 1} \right)\) (A1)
are different R1
[2 marks]
METHOD 2
for boat A, \(9t = 3 \Rightarrow t = \frac{1}{3}\) and for boat B, \(7 – 4t = 3 \Rightarrow t = 1\)
times are different so boats do not collide R1AG
[2 marks]
Question
Consider the triangle \(ABC\). The points \(P\), \(Q\) and \(R\) are the midpoints of the line segments [\(AB\)], [\(BC\)] and [\(AC\)] respectively.
Let \(\overrightarrow {{\text{OA}}} = {{a}}\), \(\overrightarrow {{\text{OB}}} = {{b}}\) and \(\overrightarrow {{\text{OC}}} = {{c}}\).
a.Find \(\overrightarrow {{\text{BR}}} \) in terms of \({{a}}\), \({{b}}\) and \({{c}}\).[2]
b.(i) Find a vector equation of the line that passes through \(B\) and \(R\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\lambda \).
(ii) Find a vector equation of the line that passes through \(A\) and \(Q\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\mu \).
(iii) Hence show that \(\overrightarrow {{\text{OG}}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) given that \(G\) is the point where [\(BR\)] and [\(AQ\)] intersect.[9]
c.Show that the line segment [\(CP\)] also includes the point \(G\).[3]
d.The coordinates of the points \(A\), \(B\) and \(C\) are \((1,{\text{ }}3,{\text{ }}1)\), \((3,{\text{ }}7,{\text{ }} – 5)\) and \((2,{\text{ }}2,{\text{ }}1)\) respectively.
A point \(X\) is such that [\(GX\)] is perpendicular to the plane \(ABC\).
Given that the tetrahedron \(ABCX\) has volume \({\text{12 unit}}{{\text{s}}^{\text{3}}}\), find possible coordinates
of \(X\).[9]
▶️Answer/Explanation
Markscheme
\(\overrightarrow {{\text{BR}}} = \overrightarrow {{\text{BA}}} + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}} + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\) (M1)
\( = ({{a}} – {{b}}) + \frac{1}{2}({{c}} – {{a}})\)
\( = \frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}\) A1
[2 marks]
(i) \({{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)\) A1A1
Note: Award A1A0 if the \({\text{r}} = \) is omitted in an otherwise correct expression/equation.
Do not penalise such an omission more than once.
(ii) \(\overrightarrow {{\text{AQ}}} = – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}\) (A1)
\({{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)\) A1
Note: Accept the use of the same parameter in (i) and (ii).
(iii) when \(\overrightarrow {{\text{AQ}}} \) and \(\overrightarrow {{\text{BP}}} \) intersect we will have \({{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}\) (M1)
Note: If the same parameters are used for both equations, award at most M1M1A0A0M1.
\(\frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}\)
attempt to equate the coefficients of the vectors \({{a}}\), \({{b}}\) and \({{c}}\) M1
\(\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 – \mu } \\ {1 – \lambda = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}\) (A1)
\(\lambda = \frac{2}{3}\) or \(\mu = \frac{2}{3}\) A1
substituting parameters back into one of the equations M1
\(\overrightarrow {{\text{OG}}} = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 – \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) AG
Note: Accept solution by verification.
[9 marks]
\(\overrightarrow {{\text{CP}}} = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}\) (M1)A1
so we have that \({{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}} \right)\) and when \(\beta = \frac{2}{3}\) the line passes through
the point \(G\) (ie, with position vector \(\frac{1}{3}({{a}} + {{b}} + {{c}})\)) R1
hence [\(AQ\)], [\(BR\)] and [\(CP\)] all intersect in \(G\) AG
[3 marks]
\(\overrightarrow {{\text{OG}}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { – 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right)\) A1
Note: This independent mark for the vector may be awarded wherever the vector is calculated.
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 6} \\ { – 6} \\ { – 6} \end{array}} \right)\) M1A1
\(\overrightarrow {{\text{GX}}} = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\) (M1)
volume of Tetrahedron given by \(\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}\)
\( = \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12\) (M1)(A1)
Note: Accept alternative methods, for example the use of a scalar triple product.
\( = \frac{1}{6}\sqrt {{{( – 6)}^2} + {{( – 6)}^2} + {{( – 6)}^2}} \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}} = 12\) (A1)
\( = \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3 = 12\)
\( \Rightarrow |\alpha | = 4\) A1
Note: Condone absence of absolute value.
this gives us the position of \(X\) as \(\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)\)
\({\text{X}}(6,{\text{ }}8,{\text{ }}3)\) or \(( – 2,{\text{ }}0,{\text{ }} – 5)\) A1
Note: Award A1 for either result.
[9 marks]
Total [23 marks]