IBDP Maths analysis and approaches Topic: AHL 3.12 :Algebraic and geometric approaches to the sum and difference of two vectors HL Paper 1

Question

The three vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\) and \(\boldsymbol{c}\) are given by\[{\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}
  {2y} \\
  { – 3x} \\
  {2x}
\end{array}} \right),{\text{ }}{\boldsymbol{b}}{\text{ }} = \left( {\begin{array}{*{20}{c}}
  {4x} \\
  y \\
  {3 – x}
\end{array}} \right),{\text{ }}{\boldsymbol{c}}{\text{ }} = \left( {\begin{array}{*{20}{c}}
  4 \\
  { – 7} \\
  6
\end{array}} \right){\text{ where }}x,y \in \mathbb{R}{\text{ }}{\text{.}}\]

(a)     If a + 2bc = 0, find the value of x and of y.

(b)     Find the exact value of \(|\)a + 2b\(|\).

▶️Answer/Explanation

Markscheme

(a)     \(2y + 8x = 4\)     M1

\( – 3x + 2y = – 7\)     A1

\(2x + 6 – 2x = 6\)

Note: Award M1 for attempt at components, A1 for two correct equations.

   No penalty for not checking the third equation.

 

solving : x = 1, y = –2     A1

 

(b)     \(|\)a + 2b\(| = \left| {\left( {\begin{array}{*{20}{c}}
  { – 4} \\
  { – 3} \\
  2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
  4 \\
  { – 2} \\
  2
\end{array}} \right)} \right|\)

\( = \left| {\left( {\begin{array}{*{20}{c}}
  4 \\
  { – 7} \\
  6
\end{array}} \right)} \right|\)

\( \Rightarrow |\)a + 2b\(|\) \( = \sqrt {{4^2} + {{( – 7)}^2} + {6^2}} \)     (M1)

\( = \sqrt {101} \)     A1  [5 marks]

 

Question

Let \(\alpha \) be the angle between the unit vectors a and b, where \(0 \leqslant \alpha \leqslant \pi \).

(a)     Express \(|\)ab\(|\) and \(|\)a + b\(|\) in terms of \(\alpha \).

(b)     Hence determine the value of \(\cos \alpha \) for which \(|\)a + b\(|\) = 3 \(|\)ab\(|\).

▶️Answer/Explanation

Markscheme

METHOD 1

(a)     \(|\)ab\(|\) = \(\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} – 2\left| a \right|\left| b \right|\cos \alpha } \)     M1

\( = \sqrt {2 – 2\cos \alpha } \)     A1

\(|\)a + b\(|\) = \(\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} – 2\left| a \right|\left| b \right|\cos (\pi – \alpha )} \)

\( = \sqrt {2 + 2\cos \alpha } \)     A1

Note: Accept the use of a, b for \(|\)a\(|\), \(|\)b\(|\).

 

(b)     \( = \sqrt {2 + 2\cos \alpha } = 3\sqrt {2 – 2\cos \alpha } \)     M1

\(\cos \alpha = \frac{4}{5}\)     A1 

METHOD 2

(a)     \(|\)ab\(|\) = \(2\sin \frac{\alpha}{2}\)     M1A1

\(|\)a + b\(|\) = \(2\sin \left( {\frac{\pi }{2} – \frac{\alpha}{2}} \right) = 2\cos \frac{\alpha}{2}\)     A1

Note: Accept the use of a, b for \(|\)a \(|\), \(|\)b\(|\).

 

(b)     \(2\cos \frac{\alpha}{2} = 6\sin \frac{\alpha }{2}\)

\(\tan \frac{\alpha }{2} = \frac{1}{3} \Rightarrow {\cos ^2}\frac{\alpha }{2} = \frac{9}{{10}}\)     M1

\(\cos \alpha = 2{\cos ^2}\frac{\alpha }{2} – 1 = \frac{4}{5}\)     A1

[5 marks]

 

Question

In the diagram below, [AB] is a diameter of the circle with centre O. Point C is on the circumference of the circle. Let \(\overrightarrow {{\text{OB}}} = \boldsymbol{b} \) and \(\overrightarrow {{\text{OC}}} = \boldsymbol{c}\) .

 

a. Find an expression for \(\overrightarrow {{\text{CB}}} \) and for \(\overrightarrow {{\text{AC}}} \) in terms of \(\boldsymbol{b}\) and \(\boldsymbol{c}\) . [2]

 

b. Hence prove that \({\rm{A\hat CB}}\) is a right angle. [3]

 
▶️Answer/Explanation

Markscheme

a.

\(\overrightarrow {{\text{CB}}} = \boldsymbol{b} – \boldsymbol{c}\) , \(\overrightarrow {{\text{AC}}} = \boldsymbol{b} + \boldsymbol{c}\)     A1A1

Note: Condone absence of vector notation in (a).

[2 marks]

b.

\(\overrightarrow {{\text{AC}}}  \cdot \overrightarrow {{\text{CB}}}  = \)(b + c)\( \cdot \)(bc)     M1

= \(|\)b\({|^2}\) – \(|\)c\({|^2}\)     A1

= 0 since \(|\)b\(|\) = \(|\)c\(|\)     R1

Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.

 

so \(\overrightarrow {{\text{AC}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) i.e. \({\rm{A\hat CB}}\) is a right angle     AG

[3 marks]

 
 

Question

a. Show that the points \({\text{O}}(0,{\text{ }}0,{\text{ }}0)\), \({\text{ A}}(6,{\text{ }}0,{\text{ }}0)\), \({\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })\), \({\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})\) form a square.[3]

 

b.Find the coordinates of M, the mid-point of [OB].[1]

 

c.Show that an equation of the plane \({\mathit{\Pi }}\), containing the square OABC, is \(y + \sqrt 2 z = 0\).[3]

 

d.Find a vector equation of the line \(L\), through M, perpendicular to the plane \({\mathit{\Pi }}\).[3]

 

e.Find the coordinates of D, the point of intersection of the line \(L\) with the plane whose equation is \(y = 0\).[3]

 

f.Find the coordinates of E, the reflection of the point D in the plane \({\mathit{\Pi }}\).[3]

 

g.(i)     Find the angle \({\rm{O\hat DA}}\).

(ii)     State what this tells you about the solid OABCDE. [6]

 
▶️Answer/Explanation

Markscheme

a.

\(\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6\)   (therefore a rhombus)     A1A1

Note:     Award A1 for two correct lengths, A2 for all four.

Note: Award A1A0 for \(\overrightarrow {{\rm{OA}}}  = \overrightarrow {{\rm{CB}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}}  = \overrightarrow {A{\rm{B}}}  = \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right)\) if no magnitudes are shown.

\(\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = 0 \)   (therefore a square)     A1

Note:     Other arguments are possible with a minimum of three conditions.

[3 marks]

b.

\({\text{M}}\left( {3,{\text{ }} – \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} – \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)\)     A1

[1 mark]

c.

METHOD 1

\(\overrightarrow {{\text{OA}}}  \times \overrightarrow {{\text{OC}}}  = \)\(\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ – 6\sqrt {12} \\ – 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ – 12\sqrt 3 \\ – 12\sqrt 6 \end{array} \right)} \right)\)     M1A1

Note:     Candidates may use other pairs of vectors.

equation of plane is \( – 6\sqrt {12} y – 6\sqrt {24} z = d\)

any valid method showing that \(d = 0\)     M1

\(\mathit{\Pi} :y+\sqrt{2z}=0\)     AG

 

METHOD 2

equation of plane is \(ax + by + cz = d\)

substituting O to find \(d = 0\)     (M1)

substituting two points (A, B, C or M)     M1

eg

\(6a = 0,{\text{ }} – \sqrt {24} b + \sqrt {12} c = 0\)     A1

\(\mathit{\Pi} :y+\sqrt{2z}=0\)     AG

[3 marks]

d.

\(\boldsymbol{r} = \left( \begin{array}{c}3\\ – \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)\)     A1A1A1

Note:     Award A1 for r = , A1A1 for two correct vectors.

[3 marks]

e.

Using \(y = 0\) to find \(\lambda \)     M1

Substitute their \(\lambda \) into their equation from part (d)     M1

D has coordinates \(\left( {{\text{3, 0, 3}}\sqrt 3 } \right)\)     A1

[3 marks]

f.

\(\lambda \) for point E is the negative of the \(\lambda \) for point D     (M1)

Note:     Other possible methods may be seen.

E has coordinates \(\left( {{\text{3, }} – 2\sqrt 6 ,{\text{ }} – \sqrt 3 } \right)\)     A1A1

Note:     Award A1 for each of the y and z coordinates.

[3 marks]

g.

(i)     \(\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}}  = \)\(\left( \begin{array}{c}3\\0\\ – 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} – 3\\0\\ – 3\sqrt 3 \end{array} \right) = 18\)     M1A1

\(\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}\)     M1

hence \({\rm{O\hat DA}} = 60^\circ \)     A1

Note:     Accept method showing OAD is equilateral.

(ii)     OABCDE is a regular octahedron (accept equivalent description)     A2

Note:     A2 for saying it is made up of 8 equilateral triangles

     Award A1 for two pyramids, A1 for equilateral triangles.

     (can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)

[6 marks]

 
 

Question

a.Show that the points \({\text{O}}(0,{\text{ }}0,{\text{ }}0)\), \({\text{ A}}(6,{\text{ }}0,{\text{ }}0)\), \({\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })\), \({\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})\) form a square.[3]

 

b.Find the coordinates of M, the mid-point of [OB].[1]

c.Show that an equation of the plane \({\mathit{\Pi }}\), containing the square OABC, is \(y + \sqrt 2 z = 0\).[3]

d. Find a vector equation of the line \(L\), through M, perpendicular to the plane \({\mathit{\Pi }}\).[3]

e.Find the coordinates of D, the point of intersection of the line \(L\) with the plane whose equation is \(y = 0\).[3]

f. Find the coordinates of E, the reflection of the point D in the plane \({\mathit{\Pi }}\).[3]

g. (i)     Find the angle \({\rm{O\hat DA}}\).

(ii)     State what this tells you about the solid OABCDE. [6]

 
▶️Answer/Explanation

Markscheme

a.

\(\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6\)   (therefore a rhombus)     A1A1

Note:     Award A1 for two correct lengths, A2 for all four.

Note: Award A1A0 for \(\overrightarrow {{\rm{OA}}}  = \overrightarrow {{\rm{CB}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}}  = \overrightarrow {A{\rm{B}}}  = \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right)\) if no magnitudes are shown.

\(\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}}  = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = 0 \)   (therefore a square)     A1

Note:     Other arguments are possible with a minimum of three conditions.

[3 marks]

b.

\({\text{M}}\left( {3,{\text{ }} – \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} – \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)\)     A1

[1 mark]

c

METHOD 1

\(\overrightarrow {{\text{OA}}}  \times \overrightarrow {{\text{OC}}}  = \)\(\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ – 6\sqrt {12} \\ – 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ – 12\sqrt 3 \\ – 12\sqrt 6 \end{array} \right)} \right)\)     M1A1

Note:     Candidates may use other pairs of vectors.

equation of plane is \( – 6\sqrt {12} y – 6\sqrt {24} z = d\)

any valid method showing that \(d = 0\)     M1

\(\mathit{\Pi} :y+\sqrt{2z}=0\)     AG

 

METHOD 2

equation of plane is \(ax + by + cz = d\)

substituting O to find \(d = 0\)     (M1)

substituting two points (A, B, C or M)     M1

eg

\(6a = 0,{\text{ }} – \sqrt {24} b + \sqrt {12} c = 0\)     A1

\(\mathit{\Pi} :y+\sqrt{2z}=0\)     AG

[3 marks]

d.

\(\boldsymbol{r} = \left( \begin{array}{c}3\\ – \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)\)     A1A1A1

Note:     Award A1 for r = , A1A1 for two correct vectors.

[3 marks]

e.

Using \(y = 0\) to find \(\lambda \)     M1

Substitute their \(\lambda \) into their equation from part (d)     M1

D has coordinates \(\left( {{\text{3, 0, 3}}\sqrt 3 } \right)\)     A1

[3 marks]

f.

\(\lambda \) for point E is the negative of the \(\lambda \) for point D     (M1)

Note:     Other possible methods may be seen.

E has coordinates \(\left( {{\text{3, }} – 2\sqrt 6 ,{\text{ }} – \sqrt 3 } \right)\)     A1A1

Note:     Award A1 for each of the y and z coordinates.

[3 marks]

g.

(i)     \(\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}}  = \)\(\left( \begin{array}{c}3\\0\\ – 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} – 3\\0\\ – 3\sqrt 3 \end{array} \right) = 18\)     M1A1

\(\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}\)     M1

hence \({\rm{O\hat DA}} = 60^\circ \)     A1

Note:     Accept method showing OAD is equilateral.

(ii)     OABCDE is a regular octahedron (accept equivalent description)     A2

Note:     A2 for saying it is made up of 8 equilateral triangles

     Award A1 for two pyramids, A1 for equilateral triangles.

     (can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)

[6 marks]

 

 

 
 
 
 

Question

PQRS is a rhombus. Given that \(\overrightarrow {{\text{PQ}}}  = \) \(\boldsymbol{a}\) and \(\overrightarrow {{\text{QR}}}  = \) \(\boldsymbol{b}\),

(a)     express the vectors \(\overrightarrow {{\text{PR}}} \) and \(\overrightarrow {{\text{QS}}} \) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\);

(b)     hence show that the diagonals in a rhombus intersect at right angles.

Answer/Explanation

Markscheme

(a)     \(\overrightarrow {{\text{PR}}}  = \) a + b     A1

\(\overrightarrow {{\text{QS}}}  = \) ba     A1

[2 marks]

 

(b)     \(\overrightarrow {{\text{PR}}}  \cdot \overrightarrow {{\text{QS}}}  = \) (a + b) \( \cdot \) (a)     M1

\( = |\)b\({|^2} – |\)a\({|^2}\)     A1

for a rhombus \(|\)a\(| = |\)b\(|\)     R1

hence \(|\)b\({|^2} – |\)a\({|^2} = 0\)     A1

Note:     Do not award the final A1 unless R1 is awarded.

hence the diagonals intersect at right angles     AG

[4 marks] Total [6 marks]

 

 

Question

Consider the triangle \(ABC\). The points \(P\), \(Q\) and \(R\) are the midpoints of the line segments [\(AB\)], [\(BC\)] and [\(AC\)] respectively.

Let \(\overrightarrow {{\text{OA}}}  = {{a}}\), \(\overrightarrow {{\text{OB}}}  = {{b}}\) and \(\overrightarrow {{\text{OC}}}  = {{c}}\).

a. Find \(\overrightarrow {{\text{BR}}} \) in terms of \({{a}}\), \({{b}}\) and \({{c}}\).[2]

(i)     Find a vector equation of the line that passes through \(B\) and \(R\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\lambda \).

(ii)     Find a vector equation of the line that passes through \(A\) and \(Q\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\mu \).

(iii)     Hence show that \(\overrightarrow {{\text{OG}}}  = \frac{1}{3}({{a}} + {{b}} + {{c}})\) given that \(G\) is the point where [\(BR\)] and [\(AQ\)] intersect. [9]

b. Show that the line segment [\(CP\)] also includes the point \(G\). [3]
c. The coordinates of the points \(A\)\(B\) and \(C\) are \((1,{\text{ }}3,{\text{ }}1)\), \((3,{\text{ }}7,{\text{ }} – 5)\) and \((2,{\text{ }}2,{\text{ }}1)\) respectively.

d. A point \(X\) is such that [\(GX\)] is perpendicular to the plane \(ABC\).

Given that the tetrahedron \(ABCX\) has volume \({\text{12 unit}}{{\text{s}}^{\text{3}}}\), find possible coordinates

of \(X\).[9]

 
▶️Answer/Explanation

Markscheme

a.

\(\overrightarrow {{\text{BR}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}}  + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\)     (M1)

\( = ({{a}} – {{b}}) + \frac{1}{2}({{c}} – {{a}})\)

\( = \frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}\)     A1

[2 marks]

b.

(i)     \({{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} – {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)\)     A1A1

Note:     Award A1A0 if the \({\text{r}} = \) is omitted in an otherwise correct expression/equation.

Do not penalise such an omission more than once.

(ii)     \(\overrightarrow {{\text{AQ}}}  =  – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}\)     (A1)

\({{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { – {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)\)     A1

Note:     Accept the use of the same parameter in (i) and (ii).

(iii)     when \(\overrightarrow {{\text{AQ}}} \) and \(\overrightarrow {{\text{BP}}} \) intersect we will have \({{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}\)     (M1)

Note:     If the same parameters are used for both equations, award at most M1M1A0A0M1.

\(\frac{\lambda }{2}{{a}} + (1 – \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 – \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}\)

attempt to equate the coefficients of the vectors \({{a}}\), \({{b}}\) and \({{c}}\)     M1

\(\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 – \mu } \\ {1 – \lambda  = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}\)     (A1)

\(\lambda  = \frac{2}{3}\) or \(\mu  = \frac{2}{3}\)     A1

substituting parameters back into one of the equations     M1

\(\overrightarrow {{\text{OG}}}  = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 – \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\)     AG

Note:     Accept solution by verification.

[9 marks]

c.

\(\overrightarrow {{\text{CP}}}  = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}\)     (M1)A1

so we have that \({{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} – {{c}}} \right)\) and when \(\beta  = \frac{2}{3}\) the line passes through

the point \(G\) (ie, with position vector \(\frac{1}{3}({{a}} + {{b}} + {{c}})\))     R1

hence [\(AQ\)], [\(BR\)] and [\(CP\)] all intersect in \(G\)     AG

[3 marks]

d.

\(\overrightarrow {{\text{OG}}}  = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { – 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right)\)     A1

Note:     This independent mark for the vector may be awarded wherever the vector is calculated.

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 6} \\ { – 6} \\ { – 6} \end{array}} \right)\)     M1A1

\(\overrightarrow {{\text{GX}}}  = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\)     (M1)

volume of Tetrahedron given by \(\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}\)

\( = \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12\)     (M1)(A1)

Note:     Accept alternative methods, for example the use of a scalar triple product.

\( = \frac{1}{6}\sqrt {{{( – 6)}^2} + {{( – 6)}^2} + {{( – 6)}^2}}  \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}}  = 12\)     (A1)

\( = \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3  = 12\)

\( \Rightarrow |\alpha | = 4\)     A1

Note:     Condone absence of absolute value.

this gives us the position of \(X\) as \(\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)\)

\({\text{X}}(6,{\text{ }}8,{\text{ }}3)\) or \(( – 2,{\text{ }}0,{\text{ }} – 5)\)     A1

Note:     Award A1 for either result.

[9 marks]

Total [23 marks]

 

 

 
 
 

Question

O, A, B and C are distinct points such that \(\overrightarrow {{\text{OA}}}  = \) a, \(\overrightarrow {{\text{OB}}}  = \) b and \(\overrightarrow {{\text{OC}}}  = \) c.

It is given that c is perpendicular to \(\overrightarrow {{\text{AB}}} \) and b is perpendicular to \(\overrightarrow {{\text{AC}}} \).

Prove that a is perpendicular to \(\overrightarrow {{\text{BC}}} \).

▶️Answer/Explanation

Markscheme

\( \bullet \) (b \( – \) a) \( = 0\)     M1

Note:     Allow c \( \bullet \) \(\overrightarrow {{\text{AB}}}  = 0\) or similar for M1.

c \( \bullet \) b \( = \) c \( \bullet \) a     A1

b \( \bullet \) (c \( – \) a) \( = 0\)

b \( \bullet \) c \( = \) b \( \bullet \) a     A1

c \( \bullet \) a \( = \) b \( \bullet \) a     M1

(c \( – \) b) \( \bullet \) a \( = 0\)     A1

hence a is perpendicular to \(\overrightarrow {{\text{BC}}} \)     AG

Note:     Only award the final A1 if a dot is used throughout to indicate scalar product.

Condone any lack of specific indication that the letters represent vectors. [5 marks]

 

Question

In the following diagram, \(\overrightarrow {{\text{OA}}} \) = a, \(\overrightarrow {{\text{OB}}} \) = b. C is the midpoint of [OA] and \(\overrightarrow {{\text{OF}}}  = \frac{1}{6}\overrightarrow {{\text{FB}}} \).

It is given also that \(\overrightarrow {{\text{AD}}}  = \lambda \overrightarrow {{\text{AF}}} \) and \(\overrightarrow {{\text{CD}}}  = \mu \overrightarrow {{\text{CB}}} \), where \(\lambda ,{\text{ }}\mu  \in \mathbb{R}\).

a.i. Find, in terms of a and \(\overrightarrow {{\text{OF}}} \).[1]

a.ii.Find, in terms of a and \(\overrightarrow {{\text{AF}}} \).[2]

b.i.Find an expression for \(\overrightarrow {{\text{OD}}} \) in terms of a, b and \(\lambda \);[2]

b.ii.Find an expression for \(\overrightarrow {{\text{OD}}} \) in terms of a, b and \(\mu \).[2]

c.Show that \(\mu  = \frac{1}{{13}}\), and find the value of \(\lambda \).[4]

d.Deduce an expression for \(\overrightarrow {{\text{CD}}} \) in terms of a and b only.[2]

e.Given that area \(\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})\), find the value of \(k\).[5]
 
▶️Answer/Explanation

Markscheme

a.i.\(\overrightarrow {{\text{OF}}}  = \frac{1}{7}\)b     A1

[1 mark]

a.ii.

\(\overrightarrow {{\text{AF}}}  = \overrightarrow {{\text{OF}}}  – \overrightarrow {{\text{OA}}} \)     (M1)

\( = \frac{1}{7}\)ba     A1

[2 marks]

b.i.

\(\overrightarrow {{\text{OD}}}  = \) a \( + \lambda \left( {\frac{1}{7}b -a} \right){\text{ }}\left( { = (1 – \lambda )a + \frac{\lambda }{7}b} \right)\)     M1A1

[2 marks]

b.ii.

\(\overrightarrow {{\text{OD}}}  = \frac{1}{2}\) a \( + \mu \left( { – \frac{1}{2}a + b} \right){\text{ }}\left( { = \left( {\frac{1}{2} – \frac{\mu }{2}} \right)a + \mu b} \right)\)     M1A1

[2 marks]

c.

equating coefficients:     M1

\(\frac{\lambda }{7} = \mu ,{\text{ }}1 – \lambda  = \frac{{1 – \mu }}{2}\)     A1

solving simultaneously:     M1

\(\lambda  = \frac{7}{{13}},{\text{ }}\mu  = \frac{1}{{13}}\)     A1AG

[4 marks]

d.

\(\overrightarrow {{\text{CD}}}  = \frac{1}{{13}}\overrightarrow {{\text{CB}}} \)

\( = \frac{1}{{13}}\left( {b – \frac{1}{2}a} \right){\text{ }}\left( { =  – \frac{1}{{26}}a + \frac{1}{{13}}b} \right)\)     M1A1

[2 marks]

e.

METHOD 1

\({\text{area }}\Delta {\text{ACD}} = \frac{1}{2}{\text{CD}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}\)     (M1)

\({\text{area }}\Delta {\text{ACB}} = \frac{1}{2}{\text{CB}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}\)     (M1)

\({\text{ratio }}\frac{{{\text{area }}\Delta {\text{ACD}}}}{{{\text{area }}\Delta {\text{ACB}}}} = \frac{{{\text{CD}}}}{{{\text{CB}}}} = \frac{1}{{13}}\)     A1

\(k = \frac{{{\text{area }}\Delta {\text{OAB}}}}{{{\text{area }}\Delta {\text{CAD}}}} = \frac{{13}}{{{\text{area }}\Delta {\text{CAB}}}} \times {\text{area }}\Delta {\text{OAB}}\)     (M1)

\( = 13 \times 2 = 26\)     A1

METHOD 2

\({\text{area }}\Delta {\text{OAB}} = \frac{1}{2}\left| {a \times b} \right|\)     A1

\({\text{area }}\Delta {\text{CAD}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}}  \times \overrightarrow {{\text{CD}}} } \right|\) or \(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}}  \times \overrightarrow {{\text{AD}}} } \right|\)     M1

\( = \frac{1}{2}\left| {\frac{1}{2}a \times \left( { – \frac{1}{{26}}a + \frac{1}{{13}}b} \right)} \right|\)

\( = \frac{1}{2}\left| {\frac{1}{2}a \times \left( { – \frac{1}{{26}}a} \right) + \frac{1}{2}a \times \frac{1}{{13}}b} \right|\)     (M1)

\( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{{13}}\left| {a \times b} \right|{\text{ }}\left( { = \frac{1}{{52}}\left| {a \times b} \right|} \right)\)     A1

\({\text{area }}\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})\)

\(\frac{1}{2}\left| {a \times b} \right| = k\frac{1}{{52}}\left| {a \times b} \right|\)

\(k = 26\)     A1 [5 marks]

Question

The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.

It is given that \(\mathop {{\text{AB}}}\limits^ \to   = \mathop {{\text{DC}}}\limits^ \to  \).

The position vectors \(\mathop {{\text{OA}}}\limits^ \to  \), \(\mathop {{\text{OB}}}\limits^ \to  \), \(\mathop {{\text{OC}}}\limits^ \to  \) and \(\mathop {{\text{OD}}}\limits^ \to  \) are given by

a = i + 2j − 3k

b = 3ij + pk

c = qi + j + 2k

d = −i + rj − 2k

where p , q and r are constants.

The point where the diagonals of ABCD intersect is denoted by M.

The plane \(\Pi \) cuts the x, y and z axes at X , Y and Z respectively.

a.i.Explain why ABCD is a parallelogram.[1]

a.ii.Using vector algebra, show that \(\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to  \).[3]

b.Show that p = 1, q = 1 and r = 4.[5]

c.Find the area of the parallelogram ABCD.[4]

d. Find the vector equation of the straight line passing through M and normal to the plane \(\Pi \) containing ABCD.[4]

e. Find the Cartesian equation of \(\Pi \).[3]

f.i. Find the coordinates of X, Y and Z.[2]

f.ii.Find YZ.[2]

▶️Answer/Explanation

Markscheme

a.i. a pair of opposite sides have equal length and are parallel      R1

hence ABCD is a parallelogram      AG

[1 mark]

a.ii.

attempt to rewrite the given information in vector form       M1

ba = cd      A1

rearranging d − a = c − b       M1

hence  \(\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to  \)     AG

Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.

[3 marks]

b.

EITHER

use of \(\mathop {{\text{AB}}}\limits^ \to   = \mathop {{\text{DC}}}\limits^ \to  \)     (M1)

\(\left( \begin{gathered}
2 \hfill \\
– 3 \hfill \\
p + 3 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q + 1 \hfill \\
1 – r \hfill \\
4 \hfill \\
\end{gathered} \right)\)       A1A1

OR

use of \(\mathop {{\text{AD}}}\limits^ \to   = \mathop {{\text{BC}}}\limits^ \to  \)      (M1)

\(\left( \begin{gathered}
– 2 \hfill \\
r – 2 \hfill \\
1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q – 3 \hfill \\
2 \hfill \\
2 – p \hfill \\
\end{gathered} \right)\)      A1A1

THEN

attempt to compare coefficients of i, j, and k in their equation or statement to that effect       M1

clear demonstration that the given values satisfy their equation       A1
p = 1, q = 1, r = 4       AG

[5 marks]

c.

attempt at computing \(\mathop {{\text{AB}}}\limits^ \to  \, \times \mathop {{\text{AD}}}\limits^ \to  \) (or equivalent)       M1

\(\left( \begin{gathered}
– 11 \hfill \\
– 10 \hfill \\
– 2 \hfill \\
\end{gathered} \right)\)     A1

area \( = \left| {\mathop {{\text{AB}}}\limits^ \to  \, \times \mathop {{\text{AD}}}\limits^ \to  } \right|\left( { = \sqrt {225} } \right)\)      (M1)

= 15       A1

[4 marks]

d.

valid attempt to find \(\mathop {{\text{OM}}}\limits^ \to   = \left( {\frac{1}{2}\left( {a + c} \right)} \right)\)      (M1)

\(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right)\)     A1

the equation is

r = \(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
11 \hfill \\
10 \hfill \\
2 \hfill \\
\end{gathered} \right)\) or equivalent       M1A1

Note: Award maximum M1A0 if ‘r = …’ (or equivalent) is not seen.

[4 marks]

e.

attempt to obtain the equation of the plane in the form ax + by + cz = d       M1

11x + 10y + 2z = 25      A1A1

Note: A1 for right hand side, A1 for left hand side.

[3 marks]

f.i.

putting two coordinates equal to zero       (M1)

\({\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)\)      A1

[2 marks]

f.ii.

\({\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}} \)     M1

\( = \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)\)     A1

[4 marks]

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