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.

$\overrightarrow {\text{AB}}  = \left( {\begin{array}{*{20}{c}}   0 \\   6 \\   { - 6} \end{array}} \right) \Rightarrow {\text{AB}} = \sqrt {72}$     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]

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.