Question
Given any two non-zero vectors a and b , show that \(|\)a \( \times \) b\({|^2}\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\).
Answer/Explanation
Ans:
METHOD 1
Use of \(|\)a \( \times \) b\(|\) = \(|\)a\(|\)\(|\)b\(|\)\(\sin \theta \) (M1)
\(|\)a \( \times \) b\({|^2}\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\sin ^2}\theta \) (A1)
Note: Only one of the first two marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\((1 – {\cos ^2}\theta )\) A1
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\cos ^2}\theta \) (A1)
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(\left( | \right.\)a\(|\)\(|\)b\(|\)\({\left. {\cos \theta } \right)^2}\) (A1)
Note: Only one of the above two A1 marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\) A1
Hence LHS = RHS AG N0
[6 marks]
METHOD 2
Use of a \( \cdot \) b = \(|\)a\(|\)\(|\)b\(|\)\(\cos \theta \) (M1)
\(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(\left( | \right.\)a\(|\)\(|\)b\(|\)\({\left. {\cos \theta } \right)^2}\) (A1)
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(|\)a\({|^2}\)\(|\)b\({|^2}\) \({\cos ^2}\theta \) (A1)
Note: Only one of the above two A1 marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\((1 – {\cos ^2}\theta )\) A1
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\sin ^2}\theta \) A1
= \(|\)a \( \times \) b\({|^2}\) A1
Hence LHS = RHS AG N0
[6 marks]
Question
Three distinct non-zero vectors are given by \(\overrightarrow {{\text{OA}}} \) = a, \(\overrightarrow {{\text{OB}}} \) = b, and \(\overrightarrow {{\text{OC}}} \) = c .
If \(\overrightarrow {{\text{OA}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) and \(\overrightarrow {{\text{OB}}} \) is perpendicular to \(\overrightarrow {{\text{CA}}} \) , show that \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \).
Answer/Explanation
Markscheme
\(\overrightarrow {{\text{BC}}} \) = c – b
\(\overrightarrow {{\text{CA}}} \) = a – c
\( \Rightarrow \) a\( \cdot \)(c – b) = 0 M1
and b\( \cdot \)(a – c) = 0 M1
\( \Rightarrow \) a\( \cdot \)c = a\( \cdot \)b A1
and a\( \cdot \)b = b\( \cdot \)c A1
\( \Rightarrow \) a\( \cdot \)c = b\( \cdot \)c M1
\( \Rightarrow \) b\( \cdot \)c – a\( \cdot \)c = 0
c\( \cdot \)(b – a) = 0 A1
\( \Rightarrow \) \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \) , as b \( \ne \) a . AG
[6 marks]
Examiners report
Only the better candidates were able to make significant progress with this question. Many candidates understood how to begin the question, but did not see how to progress to the last stage. On the whole the candidates’ use of notation in this question was poor.
Question
Consider the vectors \(\overrightarrow {{\text{OA}}} \) = a, \(\overrightarrow {{\text{OB}}} \) = b and \(\overrightarrow {{\text{OC}}} \) = a + b. Show that if \(|\)a\(|\) = \(|\)b\(|\) then (a + b)\( \cdot \)(a − b) = 0. Comment on what this tells us about the parallelogram OACB.
Answer/Explanation
Markscheme
(a + b)\( \cdot \)(a – b) = a\( \cdot \)a + b\( \cdot \)a – a\( \cdot \)b – b\( \cdot \)b M1
= a\( \cdot \)a – b\( \cdot \)b A1
= \(|\)a\({|^2}\) – \(|\)b\({|^2}\) = 0 since \(|\)a\(|\) = \(|\)b\(|\) A1
the diagonals are perpendicular R1
Note: Accept geometric proof, awarding M1 for recognizing OACB is a rhombus, R1 for a clear indication that (a + b) and (a – b) are the diagonals, A1 for stating that diagonals cross at right angles and A1 for “hence dot product is zero”.
Accept solutions using components in 2 or 3 dimensions.
[4 marks]
Examiners report
Many candidates found this more abstract question difficult. While there were some correct statements, they could not “show” the result that was asked. Some treated the vectors as scalars and notation was poor, making it difficult to follow what they were trying to do. Very few candidates realized that a – b and a + b were the diagonals of the parallelogram which prevented them from identifying the significance of the result proved. A number of candidates were clearly not aware of the difference between scalars and vectors.
Question
The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).
(i) Find the lengths of the sides of the triangle.
(ii) Find \(\cos {\rm{B\hat AC}}\).[6]
(i) Show that \(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \) −7i − 3j − 16k.
(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt {314} \).[5]
Find the Cartesian equation of the plane containing the triangle ABC.[3]
Find a vector equation of (AB).[2]
The point D on (AB) is such that \(\overrightarrow {{\text{OD}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) where O is the origin.
(i) Find the coordinates of D.
(ii) Show that D does not lie between A and B.[5]
Answer/Explanation
Markscheme
(i) \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OA}}} = \) 5i – j – 2k (or in column vector form) (A1)
Note: Award A1 if any one of the vectors, or its negative, representing the sides of the triangle is seen.
\(\overrightarrow {{\text{AB}}} = \) |5i – j – 2k|= \(\sqrt {30} \)
\(\overrightarrow {{\text{BC}}} = \) |–i – 3j + k|= \(\sqrt {11} \)
\(\overrightarrow {{\text{CA}}} = \) |–4i + 4j + k|= \(\sqrt {33} \) A2
Note: Award A1 for two correct and A0 for one correct.
(ii) METHOD 1
\(\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}\) M1A1
Note: Award M1 for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, A1 for the correct computation using their vectors.
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
Note: Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.
METHOD 2
using the cosine rule
\(\cos {\text{BAC}} = \frac{{30 + 33 – 11}}{{2\sqrt {30} \sqrt {33} }}\) M1A1
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
[6 marks]
\(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 3}&1 \\
{ – 4}&4&1
\end{array}} \right|\) A1
\( = \left( {( – 3) \times 1 – 1 \times 4} \right)\)i + \(\left( {1 \times ( – 4) – ( – 1) \times 1} \right)\)j + \(\left( {( – 1) \times 4 – ( – 3) \times ( – 4)} \right)\)k M1A1
= –7i – 3j – 16k AG
(ii) the area of \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} } \right|\) (M1)
\(\frac{1}{2}\sqrt {{{( – 7)}^2} + {{( – 3)}^2} + {{( – 16)}^2}} \) A1
\( = \frac{1}{2}\sqrt {314} \) AG
[5 marks]
attempt at the use of “(r – a)\( \cdot \)n = 0” (M1)
using r = xi + yj + zk, a = \(\overrightarrow {{\text{OA}}} \) and n = –7i – 3j – 16k (A1)
\(7x + 3y + 16z = 47\) A1
Note: Candidates who adopt a 2-parameter approach should be awarded, A1 for correct 2-parameter equations for x, y and z; M1 for a serious attempt at elimination of the parameters; A1 for the final Cartesian equation.
[3 marks]
r = \(\overrightarrow {{\text{OA}}} + t\overrightarrow {{\text{AB}}} \) (or equivalent) M1
r = (–i + 2j + 3k) + t (5i – j – 2k) A1
Note: Award M1A0 if “r =” is missing.
Note: Accept forms of the equation starting with B or with the direction reversed.
[2 marks]
(i) \(\overrightarrow {{\text{OD}}} = \) (–i + 2j + 3k) + t(5i – j – 2k)
statement that \(\overrightarrow {{\text{OD}}} \cdot \overrightarrow {{\text{BC}}} = 0\) (M1)
\(\left( {\begin{array}{*{20}{c}}
{ – 1 + 5t} \\
{2 – t} \\
{3 – 2t}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 3} \\
1
\end{array}} \right) = 0\) A1
\( – 2 – 4t = 0{\text{ or }}t = – \frac{1}{2}\) A1
coordinates of D are \(\left( { – \frac{7}{2},\frac{5}{2},4} \right)\) A1
Note: Different forms of \(\overrightarrow {{\text{OD}}} \) give different values of t, but the same final answer.
(ii) \(t < 0 \Rightarrow \) D is not between A and B R1
[5 marks]
Question
Two planes have equations
\[{\Pi _1}:{\text{ }}4x + y + z = 8{\text{ and }}{\Pi _2}:{\text{ }}4x + 3y – z = 0\]
Let \(L\) be the line of intersection of the two planes.
B is the point on \({\Pi _1}\) with coordinates \((a,{\text{ }}b,{\text{ }}1)\).
The point P lies on \(L\) and \({\rm{A\hat BP}} = 45^\circ \).
Find the cosine of the angle between the two planes in the form \(\sqrt {\frac{p}{q}} \) where \(p,{\text{ }}q \in \mathbb{Z}\).[4]
(i) Show that \(L\) has direction \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\).
(ii) Show that the point \({\text{A }}(1,{\text{ }}0,{\text{ }}4)\) lies on both planes.
(iii) Write down a vector equation of \(L\).[6]
Given the vector \(\overrightarrow {{\text{AB}}} \) is perpendicular to \(L\) find the value of \(a\) and the value of \(b\).[5]
Show that \({\text{AB}} = 3\sqrt 2 \).[1]
Find the coordinates of the two possible positions of \(P\).[5]
Answer/Explanation
Markscheme
Note: Throughout the question condone vectors written horizontally.
angle between planes is equal to the angles between the normal to the planes (M1)
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = 18\) (A1)
let \(\theta \) be the angle between the normal to the planes
\(\cos \theta = \frac{{18}}{{\sqrt {18} \sqrt {26} }} = \sqrt {\frac{{18}}{{26}}} {\text{ }}\left( {{\text{or equivalent, for example }}\sqrt {\frac{{324}}{{468}}} {\text{ or }}\sqrt {\frac{9}{{13}}} } \right)\) M1A1
[4 marks]
Note: Throughout the question condone vectors written horizontally.
(i) METHOD 1
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right)\) M1A1
which is a multiple of \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) R1AG
Note: Allow any equivalent wording or \(\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = 4\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\), do not allow \(\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\).
METHOD 2
let \(z = t\) (or equivalent)
solve simultaneously to get M1
\(y = t – 4,{\text{ }}x = 3 – 0.5t\) A1
hence direction vector is \(\left( {\begin{array}{*{20}{c}} { – 0.5} \\ 1 \\ 1 \end{array}} \right)\)
which is a multiple of \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) R1AG
METHOD 3
\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 2 + 2 = 0\) M1A1
\(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 6 – 2 = 0\) A1
Note: If only one scalar product is found award M0A0A0.
(ii) \({\Pi _1}:{\text{ }}4 + 0 + 4 = 8\) and \({\Pi _2}:{\text{ }}4 + 0 – 4 = 0\) R1
(iii) r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) A1A1
Note: A1 for “r \( = \)” and a correct point on the line, A1 for a parameter and a correct direction vector.
[6 marks]
Note: Throughout the question condone vectors written horizontally.
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right)\) (A1)
\(\left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0\) M1
Note: Award M0 for \(\left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0\).
\( – a + 1 + 2b – 6 = 0 \Rightarrow a – 2b = – 5\) A1
lies on \({\Pi _1}\) so \(4a + b + 1 = 8 \Rightarrow 4a + b = 7\) M1
\(a = 1,{\text{ }}b = 3\) A1
[5 marks]
Note: Throughout the question condone vectors written horizontally.
\({\text{AB}} = \sqrt {{0^2} + {3^2} + {{( – 3)}^2}} = 3\sqrt 2 \) M1AG
[1 mark]
Note: Throughout the question condone vectors written horizontally.
METHOD 1
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{AP}}} } \right| = 3\sqrt 2 \) (M1)
\(\overrightarrow {{\text{AP}}} = t\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)\) (A1)
\(\left| {3t} \right| = 3\sqrt 2 \Rightarrow t = \pm \sqrt 2 \) (M1)A1
\({\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)\) A1
[5 marks]
METHOD 2
let P have coordinates \((1 – \lambda ,{\text{ }}2\lambda ,{\text{ }}4 + 2\lambda )\) M1
\(\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 3 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BP}}} = \left( {\begin{array}{*{20}{c}} { – \lambda } \\ {2\lambda – 3} \\ {3 + 2\lambda } \end{array}} \right)\) A1
\(\cos 45^\circ = \frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} }}{{\left| {{\text{BA}}} \right|\left| {{\text{BP}}} \right|}}\) M1
Note: Award M1 even if AB rather than BA is used in the scalar product.
\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} = 18\)
\(\frac{1}{{\sqrt 2 }} = \frac{{18}}{{\sqrt {18} \sqrt {9{\lambda ^2} + 18} }}\)
\(\lambda = \pm \sqrt 2 \) A1
\({\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) and \(\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)\) A1
Note: Accept answers given as position vectors.
[5 marks]