Question
The vectors a and b are such that a \( = (3\cos \theta + 6)\)i \( + 7\) j and b \( = (\cos \theta – 2)\)i \( + (1 + \sin \theta )\)j.
Given that a and b are perpendicular,
a.show that \(3{\sin ^2}\theta – 7\sin \theta + 2 = 0\);[3]
b.find the smallest possible positive value of \(\theta \).[3]
▶️Answer/Explanation
Markscheme
attempting to form \((3\cos \theta + 6)(\cos \theta – 2) + 7(1 + \sin \theta ) = 0\) M1
\(3{\cos ^2}\theta – 12 + 7\sin \theta + 7 = 0\) A1
\(3\left( {1 – {{\sin }^2}\theta } \right) + 7\sin \theta – 5 = 0\) M1
\(3{\sin ^2}\theta – 7\sin \theta + 2 = 0\) AG
[3 marks]
attempting to solve algebraically (including substitution) or graphically for \(\sin \theta \) (M1)
\(\sin \theta = \frac{1}{3}\) (A1)
\(\theta = 0.340{\text{ }}( = 19.5^\circ )\) A1
[3 marks]
Examiners report
Part (a) was very well done. Most candidates were able to use the scalar product and \({\cos ^2}\theta = 1 – {\sin ^2}\theta \) to show the required result.
Part (b) was reasonably well done. A few candidates confused ‘smallest possible positive value’ with a minimum function value. Some candidates gave \(\theta = 0.34\) as their final answer.
Question
The points A, B and C have the following position vectors with respect to an origin O.
\(\overrightarrow {{\rm{OA}}} = 2\)i + j – 2k
\(\overrightarrow {{\rm{OB}}} = 2\)i – j + 2k
\(\overrightarrow {{\rm{OC}}} = \) i + 3j + 3k
The plane Π\(_2\) contains the points O, A and B and the plane Π\(_3\) contains the points O, A and C.
a.Find the vector equation of the line (BC).[3]
b.Determine whether or not the lines (OA) and (BC) intersect.[6]
c.Find the Cartesian equation of the plane Π\(_1\), which passes through C and is perpendicular to \(\overrightarrow {{\rm{OA}}} \).[3]
d.Show that the line (BC) lies in the plane Π\(_1\).[2]
e.Verify that 2j + k is perpendicular to the plane Π\(_2\).[3]
f.Find a vector perpendicular to the plane Π\(_3\).[1]
g.Find the acute angle between the planes Π\(_2\) and Π\(_3\).[4]
▶️Answer/Explanation
Markscheme
\(\overrightarrow {{\rm{BC}}} \) = (i + 3j + 3k) \( – \) (2i \( – \) j + 2k) = \( – \)i + 4j + k (A1)
r = (2i \( – \) j + 2k) + \(\lambda \)(\( – \)i + 4j + k)
(or r = (i + 3j + 3k) + \(\lambda \)(\( – \)i + 4j + k) (M1)A1
Note: Do not award A1 unless r = or equivalent correct notation seen.
[3 marks]
attempt to write in parametric form using two different parameters AND equate M1
\(2\mu = 2 – \lambda \)
\(\mu = – 1 + 4\lambda \)
\( – 2\mu = 2 + \lambda \) A1
attempt to solve first pair of simultaneous equations for two parameters M1
solving first two equations gives \(\lambda = \frac{4}{9},{\text{ }}\mu = \frac{7}{9}\) (A1)
substitution of these two values in third equation (M1)
since the values do not fit, the lines do not intersect R1
Note: Candidates may note that adding the first and third equations immediately leads to a contradiction and hence they can immediately deduce that the lines do not intersect.
[6 marks]
METHOD 1
plane is of the form r \( \bullet \) (2i + j \( – \) 2k) = d (A1)
d = (i + 3j + 3k) \( \bullet \) (2i + j \( – \) 2k) = \( – \)1 (M1)
hence Cartesian form of plane is \(2x + y – 2z = – 1\) A1
METHOD 2
plane is of the form \(2x + y – 2z = d\) (A1)
substituting \((1,{\text{ }}3,{\text{ }}3)\) (to find gives \(2 + 3 – 6 = – 1\)) (M1)
hence Cartesian form of plane is \(2x + y – 2z = – 1\) A1
[3 marks]
METHOD 1
attempt scalar product of direction vector BC with normal to plane M1
(\( – \)i + 4j + k) \( \bullet \) (2i + j \( – \) 2k) \( = – 2 + 4 – 2\)
\( = 0\) A1
hence BC lies in Π\(_1\) AG
METHOD 2
substitute eqn of line into plane M1
\({\text{line }}r = \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 2 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \\ 1 \end{array}} \right).{\text{ Plane }}{\pi _1}:2x + y – 2z = – 1\)
\(2(2 – \lambda ) + ( – 1 + 4\lambda ) – 2(2 + \lambda )\)
\( = – 1\) A1
hence BC lies in Π\(_1\) AG
Note: Candidates may also just substitute \(2i – j + 2k\) into the plane since they are told C lies on \({\pi _1}\).
Note: Do not award A1FT.
[2 marks]
METHOD 1
applying scalar product to \(\overrightarrow {{\rm{OA}}} \) and \(\overrightarrow {{\rm{OB}}} \) M1
(2j + k) \( \bullet \) (2i + j \( – \) 2k) = 0 A1
(2j + k) \( \bullet \) (2i \( – \) j + 2k) =0 A1
METHOD 2
attempt to find cross product of \(\overrightarrow {{\rm{OA}}} \) and \(\overrightarrow {{\rm{OB}}} \) M1
plane Π\(_2\) has normal \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \) = \( – \) 8j \( – \) 4k A1
since \( – \)8j \( – \) 4k = \( – \)4(2j + k), 2j + k is perpendicular to the plane Π\(_2\) R1
[3 marks]
plane Π\(_3\) has normal \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} \) = 9i \( – \) 8j + 5k A1
[1 mark]
attempt to use dot product of normal vectors (M1)
\(\cos \theta = \frac{{(2j + k) \bullet (9i – 8j + 5k)}}{{\left| {2j + k} \right|\left| {9i – 8j + 5k} \right|}}\) (M1)
\( = \frac{{ – 11}}{{\sqrt 5 \sqrt {170} }}\,\,\,( = – 0.377 \ldots )\) (A1)
Note: Accept \(\frac{{11}}{{\sqrt 5 \sqrt {170} }}\). acute angle between planes \( = 67.8^\circ \,\,\,{\text{(}} = 1.18^\circ )\) A1
[4 marks]
Question
The angle between the vector a = i − 2j + 3k and the vector b = 3i − 2j + mk is 30° .
Find the values of m.
▶️Answer/Explanation
Markscheme
\({\boldsymbol{a}} \cdot {\boldsymbol{b}} = \left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\cos \theta \) (M1)
\({\boldsymbol{a}} \cdot {\boldsymbol{b}} = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 2} \\
3
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
3 \\
{ – 2} \\
m
\end{array}} \right) = 7 + 3m\) A1
\(\left| {\boldsymbol{a}} \right| = \sqrt {14} \) \(\left| {\boldsymbol{b}} \right| = \sqrt {13 + {m^2}} \) A1
\(\left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\cos \theta = \sqrt {14} \sqrt {13 + {m^2}} \cos 30^\circ \)
\(7 + 3m = \sqrt {14} \sqrt {13 + {m^2}} \cos 30^\circ \) A1
m = 2.27, m = 25.7 A1A1
[6 marks]
Examiners report
Many candidates gained the first 4 marks by obtaining the equation, in unsimplified form, satisfied by m but then made mistakes in simplifying and solving this equation.
Question
Consider the vectors a \( = \sin (2\alpha )\)i \( – \cos (2\alpha )\)j + k and b \( = \cos \alpha \)i \( – \sin \alpha \)j − k, where \(0 < \alpha < 2\pi \).
Let \(\theta \) be the angle between the vectors a and b.
(a) Express \(\cos \theta \) in terms of \(\alpha \).
(b) Find the acute angle \(\alpha \) for which the two vectors are perpendicular.
(c) For \(\alpha = \frac{{7\pi }}{6}\), determine the vector product of a and b and comment on the geometrical significance of this result.
▶️Answer/Explanation
Markscheme
(a) \(\cos \theta = \frac{{\boldsymbol{ab}}}{{\left| \boldsymbol{a} \right|\left| \boldsymbol{b} \right|}} = \frac{{\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha – 1}}{{\sqrt 2 \times \sqrt 2 }}{\text{ }}\left( { = \frac{{\sin 3\alpha – 1}}{2}} \right)\) M1A1
(b) \({\boldsymbol{a}} \bot {\boldsymbol{b}} \Rightarrow \cos \theta = 0\) M1
\(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha – 1 = 0\)
\(\alpha = 0.524{\text{ }}\left( { = \frac{\pi }{6}} \right)\) A1
(c)
METHOD 1
\(\left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{\sin 2\alpha }&{ – \cos 2\alpha }&1 \\
{\cos \alpha }&{ – \sin \alpha }&{ – 1}
\end{array}} \right|\) (M1)
assuming \(\alpha = \frac{{7\pi }}{6}\)
Note: Allow substitution at any stage.
\(\left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
{\frac{{\sqrt 3 }}{2}}&{ – \frac{1}{2}}&1 \\
{ – \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}&{ – 1}
\end{array}} \right|\) A1
\(= \boldsymbol{i} \left( {\frac{1}{2} – \frac{1}{2}} \right) – \boldsymbol{j} \left( { – \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2}} \right) + \boldsymbol{k}\left( {\frac{{\sqrt 3 }}{2} \times \frac{1}{2} – \frac{1}{2} \times \frac{{\sqrt 3 }}{2}} \right)\)
= 0 A1
a and b are parallel R1
Note: Accept decimal equivalents.
METHOD 2
from (a) \(\cos \theta = – 1{\text{ (and }}\sin \theta = 0)\) M1A1
\(\boldsymbol{a} \times \boldsymbol{b}\) = 0 A1
a and b are parallel R1
[8 marks]
Examiners report
This question was attempted by most candidates who in general were able to find the dot product of the vectors in part (a). However the simplification of the expression caused difficulties which affected the performance in part (b). Many candidates had difficulties in interpreting the meaning of a \( \times \) b = 0 in part (c).
Question
Let \(u = 6i + 3j + 6k\) and \(v = 2i + 2j + k\).
a.Find
(i) \(u \bullet v\);
(ii) \(\left| {{u}} \right|\);
(iii) \(\left| {{v}} \right|\).[5]
b.Find the angle between \({{u}}\) and \({{v}}\).[2]
▶️Answer/Explanation
Markscheme
(i) correct substitution (A1)
eg\(\;\;\;6 \times 2 + 3 \times 2 + 6 \times 1\)
\(u \bullet v = 24\) A1 N2
(ii) correct substitution into magnitude formula for \({{u}}\) or \({{v}}\) (A1)
eg\(\;\;\;\sqrt {{6^2} + {3^2} + {6^2}} ,{\text{ }}\sqrt {{2^2} + {2^2} + {1^2}} \), correct value for \(\left| {{v}} \right|\)
\(\left| {{u}} \right| = 9\) A1 N2
(iii) \(\left| {{v}} \right| = 3\) A1 N1
[5 marks]
correct substitution into angle formula (A1)
eg\(\;\;\;\frac{{24}}{{9 \times 3}},{\text{ }}0.\bar 8\)
\(0.475882,{\text{ }}27.26604^\circ \) A1 N2
\(0.476,{\text{ }}27.3^\circ \)
[2 marks]
Total [7 marks]
Question
Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 3}\\
6
\end{array}} \right)\) and \({\boldsymbol{w}} = \left( {\begin{array}{*{20}{c}}
k\\
{ – 2}\\
4
\end{array}} \right)\) , for \(k > 0\) . The angle between v and w is \(\frac{\pi }{3}\) .
Find the value of \(k\) .
▶️Answer/Explanation
Markscheme
correct substitutions for \({\boldsymbol{v}} \bullet {\boldsymbol{w}}\) ; \(\left| {\boldsymbol{v}} \right|\) ; \(\left| {\boldsymbol{w}} \right|\) (A1)(A1)(A1)
e.g. \(2k + ( – 3) \times ( – 2) + 6 \times 4\) , \(2k + 30\) ; \(\sqrt {{2^2} + {{( – 3)}^2} + {6^2}} \) , \(\sqrt {49} \) ; \(\sqrt {{k^2} + {{( – 2)}^2} + {4^2}} \) , \(\sqrt {{k^2} + 20} \)
evidence of substituting into the formula for scalar product (M1)
e.g. \(\frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)
correct substitution A1
e.g. \(\cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)
\(k = 18.8\) A2 N5
[7 marks]
Question
Consider the points A(\(5\), \(2\), \(1\)) , B(\(6\), \(5\), \(3\)) , and C(\(7\), \(6\), \(a + 1\)) , \(a \in{\mathbb{R}}\) .
Let \({\rm{q}}\) be the angle between \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) .
a.Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AC}}} \) .[3]
b.Find the value of \(a\) for which \({\rm{q}} = \frac{\pi }{2}\) .[4]
c.i. Show that \(\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) .
ii. Hence, find the value of a for which \({\rm{q}} = 1.2\) .[8]
c.ii.Hence, find the value of a for which \({\rm{q}} = 1.2\) .[4]
▶️Answer/Explanation
Markscheme
(i) appropriate approach (M1)
eg \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}} \) , \({\rm{B}} – {\rm{A}}\)
\(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
1\\
3\\
2
\end{array} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
2\\
4\\
a
\end{array} \right)\) A1 N1
[3 marks]
valid reasoning (seen anywhere) R1
eg scalar product is zero, \(\cos \frac{\pi }{2} = \frac{{\boldsymbol{u} \bullet \boldsymbol{v}}}{{\left| \boldsymbol{u} \right|\left| \boldsymbol{v} \right|}}\)
correct scalar product of their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (may be seen in part (c)) (A1)
eg \(1(2) + 3(4) + 2(a)\)
correct working for their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (A1)
eg \(2a + 14\) , \(2a = – 14\)
\(a = – 7\) A1 N3
[4 marks]
correct magnitudes (may be seen in (b)) (A1)(A1)
\(\sqrt {{1^2} + {3^2} + {2^2}} \left( { = \sqrt {14} } \right)\) , \(\sqrt {{2^2} + {4^2} + {a^2}} \left( { = \sqrt {20 + {a^2}} } \right)\)
substitution into formula (M1)
eg \(\cos \theta \frac{{1 \times 2 + 3 \times 4 + 2 \times a}}{{\sqrt {{1^2} + {3^2} + {2^2}} \sqrt {{2^2} + {4^2} + {a^2}} }}\) , \(\frac{{14 + 2a}}{{\sqrt {14} \sqrt {4 + 16 + {a^2}} }}\)
simplification leading to required answer A1
eg \(\cos \theta = \frac{{14 + 2a}}{{\sqrt {14} \sqrt {20 + {a^2}} }}\)
\(\cos \theta = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) AG N0
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0\) , attempt to square
\(a = – 3.25\) A2 N3
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0\) , attempt to square
\(a = – 3.25\) A2 N3
[4 marks]
Question
Let \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)\).
a.Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).[2]
b.Let \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)\). Find \({\rm{B\hat AC}}\).[4]
▶️Answer/Explanation
Markscheme
correct substitution (A1)
eg\(\,\,\,\,\,\)\(\sqrt {{4^2} + {1^2} + {2^2}} \)
4.58257
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21} \) (exact), 4.58 A1 N2
[2 marks]
finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) (A1)(A1)
scalar product \( = (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)\)
\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + 0 + 0} {\text{ }}( = 3)\)
substituting their values into cosine formula (M1)
eg cos B\(\hat A\)C\({\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}} \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta = 0.873\)
0.509739 (29.2059°)
\({\rm{B\hat AC}} = 0.510\) (29.2°) A1 N2
[4 marks]