# IBDP Maths analysis and approaches Topic: AHL 3.12 :Algebraic and geometric approaches to multiplication by a scalar, kv HL Paper 1

## 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]

## 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]

## 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]

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.

## Markscheme

(a)     $$\overrightarrow {{\text{PR}}} =$$ ab     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]

b.(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]

c.Show that the line segment [$$CP$$] also includes the point $$G$$.[3]

d.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.

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]

## 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 $$\overrightarrow {{\text{BC}}}$$.

## 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]

## 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

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]

## 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]

## 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]

## 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]

## 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]

## 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]

## 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]

## 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]

## 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]

## 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]

## 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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