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
Given that \({\boldsymbol{a}} = 2\sin \theta {\boldsymbol{i}} + \left( {1 – \sin \theta } \right){\boldsymbol{j}}\) , find the value of the acute angle \(\theta \) , so that \(\boldsymbol{a}\) is perpendicular to the line \(x + y = 1\).
Answer/Explanation
Markscheme
direction vector for line \( = \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1}
\end{array}} \right)\) or any multiple A1
\(\left( {\begin{array}{*{20}{c}}
{2\sin \theta } \\
{1 – \sin \theta }
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
{ – 1}
\end{array}} \right) = 0\) M1
\(2\sin \theta – 1 + \sin \theta = 0\) A1
Note: Allow FT on candidate’s direction vector just for line above only.
\(3\sin \theta = 1\)
\(\sin \theta = \frac{1}{3}\) A1
\(\theta = 0.340\) or \(19.5\) A1
Note: A coordinate geometry method using perpendicular gradients is acceptable.
[5 marks]
Examiners report
A variety of approaches were seen, either using a scalar product of vectors, or based on the rule for perpendicular gradients of lines. The main problem encountered in the first approach was in the choice of the correct vector direction for the line.
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
The coordinates of points A, B and C are given as \((5,\, – 2,\,5)\) , \((5,\,4,\, – 1)\) and \(( – 1,\, – 2,\, – 1)\) respectively.
Show that AB = AC and that \({\rm{B\hat AC}} = 60^\circ \).
Find the Cartesian equation of \(\Pi \), the plane passing through A, B, and C.
(i) Find the Cartesian equation of \({\Pi _1}\) , the plane perpendicular to (AB) passing through the midpoint of [AB] .
(ii) Find the Cartesian equation of \({\Pi _2}\) , the plane perpendicular to (AC) passing through the midpoint of [AC].
Find the vector equation of L , the line of intersection of \({\Pi _1}\) and \({\Pi _2}\) , and show that it is perpendicular to \(\Pi \) .
A methane molecule consists of a carbon atom with four hydrogen atoms symmetrically placed around it in three dimensions.
The positions of the centres of three of the hydrogen atoms are A, B and C as given. The position of the centre of the fourth hydrogen atom is D.
Using the fact that \({\text{AB}} = {\text{AD}}\) , show that the coordinates of one of the possible positions of the fourth hydrogen atom is \(( -1,\,4,\,5)\) .
A methane molecule consists of a carbon atom with four hydrogen atoms symmetrically placed around it in three dimensions.
The positions of the centres of three of the hydrogen atoms are A, B and C as given. The position of the centre of the fourth hydrogen atom is D.
Letting D be \(( – 1,\,4,\,5)\) , show that the coordinates of G, the position of the centre of the carbon atom, are \((2,\,1,\,2)\) . Hence calculate \({\rm{D}}\hat {\rm{G}}{\rm{A}}\) , the bonding angle of carbon.
Answer/Explanation
Markscheme
\(\overrightarrow {\text{AB}} = \left( {\begin{array}{*{20}{c}}
0 \\
6 \\
{ – 6}
\end{array}} \right) \Rightarrow {\text{AB}} = \sqrt {72} \) A1
\(\overrightarrow {\text{AC}} = \left( {\begin{array}{*{20}{c}}
{ – 6} \\
0 \\
{ – 6}
\end{array}} \right) \Rightarrow {\text{AC}} = \sqrt {72} \) A1
so they are the same AG
\(\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{AC}}} = 36 = \left( {\sqrt {72} } \right)\left( {\sqrt {72} } \right)\cos \theta \) (M1)
\(\cos \theta = \frac{{36}}{{\left( {\sqrt {72} } \right)\left( {\sqrt {72} } \right)}} = \frac{1}{2} \Rightarrow \theta = 60^\circ \) A1AG
Note: Award M1A1 if candidates find BC and claim that triangle ABC is equilateral.
[4 marks]
METHOD 1
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\
0&6&{ – 6} \\
{ – 6}&0&{ – 6}
\end{array}} \right| = -36\boldsymbol{i} + 36\boldsymbol{j} + 36\boldsymbol{k}\) (M1)A1
equation of plane is \(x – y – z = k\) (M1)
goes through A, B or C \( \Rightarrow x – y – z = 2\) A1
[4 marks]
METHOD 2
\(x + by + cz = d\) (or similar) M1
\(5 – 2b + 5c = d\)
\(5 + 4b – c = d\) A1
\( -1 – 2b – c = d\)
solving simultaneously M1
\(b = -1,{\text{ }}c = -1,{\text{ }}d = 2\)
so \(x – y – z = 2\) A1
[4 marks]
(i) midpoint is \((5,\,1,\,2)\), so equation of \({\Pi _1}\) is \(y – z = -1\) A1A1
(ii) midpoint is \((2,\, – 2,\,2)\), so equation of \({\Pi _2}\) is \(x + z = 4\) A1A1
Note: In each part, award A1 for midpoint and A1 for the equation of the plane.
[4 marks]
EITHER
solving the two equations above M1
\(L:r = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ – 1} \\
1 \\
1
\end{array}} \right)\) A1
OR
L has the direction of the vector product of the normal vectors to the planes \({\Pi _1}\) and \({\Pi _2}\) (M1)
\(\left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
0&1&{ – 1} \\
1&0&1
\end{array}} \right| = \boldsymbol{i} – \boldsymbol{j} – \boldsymbol{k}\)
(or its opposite) A1
THEN
direction is \(\left( {\begin{array}{*{20}{c}}
{ – 1} \\
1 \\
1
\end{array}} \right)\)as required R1
[3 marks]
D is of the form \((4 – \lambda ,\, -1 + \lambda ,\,\lambda )\) M1
\({(1 + \lambda )^2} + {( -1 – \lambda )^2} + {(5 – \lambda )^2} = 72\) M1
\(3{\lambda ^2} – 6\lambda – 45 = 0\)
\(\lambda = 5{\text{ or }}\lambda = -3\) A1
\({\text{D}}( -1,\,4,\,5)\) AG
Note: Award M0M0A0 if candidates just show that \({\text{D}}( -1,\,4,\,5)\) satisfies \({\text{AB}} = {\text{AD}}\);
Award M1M1A0 if candidates also show that D is of the form \((4 – \lambda ,\, -1 + \lambda ,\,\lambda )\)
[3 marks]
EITHER
G is of the form \((4 – \lambda ,\, – 1 + \lambda ,\,\lambda )\) and \({\text{DG}} = {\text{AG, BG or CG}}\) M1
e.g. \({(1 + \lambda )^2} + {( – 1 – \lambda )^2} + {(5 – \lambda )^2} = {(5 – \lambda )^2} + {(5 – \lambda )^2} + {(5 – \lambda )^2}\) M1
\({(1 + \lambda )^2} = {(5 – \lambda )^2}\)
\(\lambda = 2\) A1
\({\text{G}}(2,\,1,\,2)\) AG
OR
G is the centre of mass (barycentre) of the regular tetrahedron ABCD (M1)
\({\text{G}}\left( {\frac{{5 + 5 + ( – 1) + ( – 1)}}{4},\frac{{ – 2 + 4 + ( – 2) + 4}}{4},\frac{{5 + ( – 1) + ( – 1) + 5}}{4}} \right)\) M1A1
THEN
Note: the following part is independent of previous work and candidates may use AG to answer it (here it is possible to award M0M0A0A1M1A1)
\(\overrightarrow {GD} = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
3 \\
3
\end{array}} \right)\) and \(\overrightarrow {GA} = \left( {\begin{array}{*{20}{c}}
3 \\
{ – 3} \\
3
\end{array}} \right)\) A1
\(\cos \theta = \frac{{ – 9}}{{\left( {3\sqrt 3 } \right)\left( {3\sqrt 3 } \right)}} = – \frac{1}{3} \Rightarrow \theta = 109^\circ \) (or 1.91 radians) M1A1
[6 marks]
Examiners report
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Question
The coordinates of points A, B and C are given as \((5,\, – 2,\,5)\) , \((5,\,4,\, – 1)\) and \(( – 1,\, – 2,\, – 1)\) respectively.
Show that AB = AC and that \({\rm{B\hat AC}} = 60^\circ \).
Find the Cartesian equation of \(\Pi \), the plane passing through A, B, and C.
(i) Find the Cartesian equation of \({\Pi _1}\) , the plane perpendicular to (AB) passing through the midpoint of [AB] .
(ii) Find the Cartesian equation of \({\Pi _2}\) , the plane perpendicular to (AC) passing through the midpoint of [AC].
Find the vector equation of L , the line of intersection of \({\Pi _1}\) and \({\Pi _2}\) , and show that it is perpendicular to \(\Pi \) .
A methane molecule consists of a carbon atom with four hydrogen atoms symmetrically placed around it in three dimensions.
The positions of the centres of three of the hydrogen atoms are A, B and C as given. The position of the centre of the fourth hydrogen atom is D.
Using the fact that \({\text{AB}} = {\text{AD}}\) , show that the coordinates of one of the possible positions of the fourth hydrogen atom is \(( -1,\,4,\,5)\) .
A methane molecule consists of a carbon atom with four hydrogen atoms symmetrically placed around it in three dimensions.
The positions of the centres of three of the hydrogen atoms are A, B and C as given. The position of the centre of the fourth hydrogen atom is D.
Letting D be \(( – 1,\,4,\,5)\) , show that the coordinates of G, the position of the centre of the carbon atom, are \((2,\,1,\,2)\) . Hence calculate \({\rm{D}}\hat {\rm{G}}{\rm{A}}\) , the bonding angle of carbon.
Answer/Explanation
Markscheme
\(\overrightarrow {\text{AB}} = \left( {\begin{array}{*{20}{c}}
0 \\
6 \\
{ – 6}
\end{array}} \right) \Rightarrow {\text{AB}} = \sqrt {72} \) A1
\(\overrightarrow {\text{AC}} = \left( {\begin{array}{*{20}{c}}
{ – 6} \\
0 \\
{ – 6}
\end{array}} \right) \Rightarrow {\text{AC}} = \sqrt {72} \) A1
so they are the same AG
\(\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{AC}}} = 36 = \left( {\sqrt {72} } \right)\left( {\sqrt {72} } \right)\cos \theta \) (M1)
\(\cos \theta = \frac{{36}}{{\left( {\sqrt {72} } \right)\left( {\sqrt {72} } \right)}} = \frac{1}{2} \Rightarrow \theta = 60^\circ \) A1AG
Note: Award M1A1 if candidates find BC and claim that triangle ABC is equilateral.
[4 marks]
METHOD 1
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\
0&6&{ – 6} \\
{ – 6}&0&{ – 6}
\end{array}} \right| = -36\boldsymbol{i} + 36\boldsymbol{j} + 36\boldsymbol{k}\) (M1)A1
equation of plane is \(x – y – z = k\) (M1)
goes through A, B or C \( \Rightarrow x – y – z = 2\) A1
[4 marks]
METHOD 2
\(x + by + cz = d\) (or similar) M1
\(5 – 2b + 5c = d\)
\(5 + 4b – c = d\) A1
\( -1 – 2b – c = d\)
solving simultaneously M1
\(b = -1,{\text{ }}c = -1,{\text{ }}d = 2\)
so \(x – y – z = 2\) A1
[4 marks]
(i) midpoint is \((5,\,1,\,2)\), so equation of \({\Pi _1}\) is \(y – z = -1\) A1A1
(ii) midpoint is \((2,\, – 2,\,2)\), so equation of \({\Pi _2}\) is \(x + z = 4\) A1A1
Note: In each part, award A1 for midpoint and A1 for the equation of the plane.
[4 marks]
EITHER
solving the two equations above M1
\(L:r = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 1} \\
0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
{ – 1} \\
1 \\
1
\end{array}} \right)\) A1
OR
L has the direction of the vector product of the normal vectors to the planes \({\Pi _1}\) and \({\Pi _2}\) (M1)
\(\left| {\begin{array}{*{20}{c}}
\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
0&1&{ – 1} \\
1&0&1
\end{array}} \right| = \boldsymbol{i} – \boldsymbol{j} – \boldsymbol{k}\)
(or its opposite) A1
THEN
direction is \(\left( {\begin{array}{*{20}{c}}
{ – 1} \\
1 \\
1
\end{array}} \right)\)as required R1
[3 marks]
D is of the form \((4 – \lambda ,\, -1 + \lambda ,\,\lambda )\) M1
\({(1 + \lambda )^2} + {( -1 – \lambda )^2} + {(5 – \lambda )^2} = 72\) M1
\(3{\lambda ^2} – 6\lambda – 45 = 0\)
\(\lambda = 5{\text{ or }}\lambda = -3\) A1
\({\text{D}}( -1,\,4,\,5)\) AG
Note: Award M0M0A0 if candidates just show that \({\text{D}}( -1,\,4,\,5)\) satisfies \({\text{AB}} = {\text{AD}}\);
Award M1M1A0 if candidates also show that D is of the form \((4 – \lambda ,\, -1 + \lambda ,\,\lambda )\)
[3 marks]
EITHER
G is of the form \((4 – \lambda ,\, – 1 + \lambda ,\,\lambda )\) and \({\text{DG}} = {\text{AG, BG or CG}}\) M1
e.g. \({(1 + \lambda )^2} + {( – 1 – \lambda )^2} + {(5 – \lambda )^2} = {(5 – \lambda )^2} + {(5 – \lambda )^2} + {(5 – \lambda )^2}\) M1
\({(1 + \lambda )^2} = {(5 – \lambda )^2}\)
\(\lambda = 2\) A1
\({\text{G}}(2,\,1,\,2)\) AG
OR
G is the centre of mass (barycentre) of the regular tetrahedron ABCD (M1)
\({\text{G}}\left( {\frac{{5 + 5 + ( – 1) + ( – 1)}}{4},\frac{{ – 2 + 4 + ( – 2) + 4}}{4},\frac{{5 + ( – 1) + ( – 1) + 5}}{4}} \right)\) M1A1
THEN
Note: the following part is independent of previous work and candidates may use AG to answer it (here it is possible to award M0M0A0A1M1A1)
\(\overrightarrow {GD} = \left( {\begin{array}{*{20}{c}}
{ – 3} \\
3 \\
3
\end{array}} \right)\) and \(\overrightarrow {GA} = \left( {\begin{array}{*{20}{c}}
3 \\
{ – 3} \\
3
\end{array}} \right)\) A1
\(\cos \theta = \frac{{ – 9}}{{\left( {3\sqrt 3 } \right)\left( {3\sqrt 3 } \right)}} = – \frac{1}{3} \Rightarrow \theta = 109^\circ \) (or 1.91 radians) M1A1
[6 marks]
Examiners report
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.
Parts (a) and (b) were well attempted with many candidates achieving full or nearly full marks in these parts. Surprisingly, many candidates were not able to find the coordinates of the midpoints in part (c) leading to incorrect equations of the planes and, in many cases this affected the performance in part (d). Very few answered part (d) correctly. In parts (e) and (f) very few candidates made good attempts of using the ‘show that’ procedure and used a parametric approach. In most attempts, candidates simply verified the condition using the answer given. In a few cases, candidates noticed that G is the barycentre of [ABCD] and found its coordinates successfully. Some candidates were also successful in determining the bounding angle using the information given. Unfortunately, many candidates instead of finding the angle simply quoted the result.