## 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

**for the equation of the plane.**

*A1**[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

**if candidates just show that \({\text{D}}( -1,\,4,\,5)\) satisfies \({\text{AB}} = {\text{AD}}\);**

*M0M0A0*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.

## 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 \(( -1,\,4,\,5)\) .

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

**for the equation of the plane.**

*A1**[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

**if candidates just show that \({\text{D}}( -1,\,4,\,5)\) satisfies \({\text{AB}} = {\text{AD}}\);**

*M0M0A0*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

## Question

The points A, B and C have the following position vectors with respect to an origin O.

\(\overrightarrow {{\rm{OA}}} = 2\)** i** +

**– 2**

*j*

*k*\(\overrightarrow {{\rm{OB}}} = 2\)** i** –

**+ 2**

*j*

*k*\(\overrightarrow {{\rm{OC}}} = \) ** i** + 3

**+ 3**

*j*

*k*The plane *Π*\(_2\) contains the points O, A and B and the plane *Π*\(_3\) contains the points O, A and C.

Find the vector equation of the line (BC).

Determine whether or not the lines (OA) and (BC) intersect.

Find the Cartesian equation of the plane *Π*\(_1\), which passes through C and is perpendicular to \(\overrightarrow {{\rm{OA}}} \).

Show that the line (BC) lies in the plane *Π*\(_1\).

Verify that 2** j **+

**is perpendicular to the plane**

*k**Π*\(_2\).

Find a vector perpendicular to the plane *Π*\(_3\).

Find the acute angle between the planes *Π*\(_2\) and *Π*\(_3\).

**Answer/Explanation**

## Markscheme

\(\overrightarrow {{\rm{BC}}} \) = (** i** + 3

**+ 3**

*j***) \( – \) (2**

*k***\( – \)**

*i***+ 2**

*j***) = \( – \)**

*k***+ 4**

*i***+**

*j*

*k*

*(A1)*** r** = (2

**\( – \)**

*i***+ 2**

*j***) + \(\lambda \)(\( – \)**

*k***+ 4**

*i***+**

*j***)**

*k*(or ** r** = (

**+ 3**

*i***+ 3**

*j***) + \(\lambda \)(\( – \)**

*k***+ 4**

*i***+**

*j***)**

*k*

*(M1)A1***Note:** Do not award ** A1 **unless

**= or equivalent correct notation seen.**

*r**[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** \( \bullet \) (2

**+**

*i***\( – \) 2**

*j***) =**

*k**d*

*(A1)**d *= (** i** + 3

**+ 3**

*j***) \( \bullet \) (2**

*k***+**

*i***\( – \) 2**

*j***) = \( – \)1**

*k*

*(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** + 4

**+**

*j***) \( \bullet \) (2**

*k***+**

*i***\( – \) 2**

*j***) \( = – 2 + 4 – 2\)**

*k*\( = 0\) *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*

(2** j** +

**) \( \bullet \) (2**

*k***+**

*i***\( – \) 2**

*j***) = 0**

*k*

*A1*(2** j** +

**) \( \bullet \) (2**

*k***\( – \)**

*i***+ 2**

*j***) =0**

*k*

*A1***METHOD 2**

attempt to find cross product of \(\overrightarrow {{\rm{OA}}} \) and \(\overrightarrow {{\rm{OB}}} \) *M1*

plane *Π*\(_2\) has normal \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \) = \( – \) 8** j **\( – \) 4

*k*

*A1*since \( – \)8** j **\( – \) 4

**= \( – \)4(2**

*k***+**

*j***), 2**

*k***+**

*j***is perpendicular to the plane**

*k**Π*\(_2\)

*R1**[3 marks]*

plane *Π*\(_3\) has normal \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} \) = 9** i** \( – \) 8

**+ 5**

*j*

*k*

*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]*

## Examiners report

[N/A]

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