Home / IBDP Maths analysis and approaches Topic: AHL 3.12 displacement vector HL Paper 1

IBDP Maths analysis and approaches Topic: AHL 3.12 displacement vector 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 \(\overrightarrow {{\text{OC}}} = \boldsymbol{c}\) .

 

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

[2]
a.

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

[3]
b.
Answer/Explanation

Markscheme

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

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

[2 marks]

a.

\(\overrightarrow {{\text{AC}}}  \cdot \overrightarrow {{\text{CB}}}  = \)(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]

b.

Examiners report

Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably. 

a.

Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably. 

b.

Question

The position vectors of the points \(A\), \(B\) and \(C\) are \(a\), \(b\) and \(c\) respectively, relative to an origin \(O\). The following diagram shows the triangle \(ABC\) and points \(M\), \(R\), \(S\) and \(T\).

\(M\) is the midpoint of [\(AC\)].

\(R\) is a point on [\(AB\)] such that \(\overrightarrow {{\text{AR}}}  = \frac{1}{3}\overrightarrow {{\text{AB}}} \).

\(S\) is a point on [\(AC\)] such that \(\overrightarrow {{\text{AS}}}  = \frac{2}{3}\overrightarrow {{\text{AC}}} \).

\(T\) is a point on [\(RS\)] such that \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \).

(i)     Express \(\overrightarrow {{\text{AM}}} \) in terms of \(a\) and \(c\).

(ii)     Hence show that \(\overrightarrow {{\text{BM}}}  = \frac{1}{2}\)\(a\) – \(b\)\( + \frac{1}{2}c\).

[4]
a.

(i)     Express \(\overrightarrow {{\text{RA}}} \) in terms of \(a\) and \(b\).

(ii)     Show that \(\overrightarrow {RT}  =  – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\).

[5]
b.

Prove that \(T\) lies on [\(BM\)].

[5]
c.
Answer/Explanation

Markscheme

(i)     \(\overrightarrow {{\text{AM}}}  = \frac{1}{2}\overrightarrow {{\text{AC}}} \)     (M1)

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

(ii)     \(\overrightarrow {{\text{BM}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AM}}} \)     M1

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

\(\overrightarrow {{\text{BM}}}  = \frac{1}{2}a – b + \frac{1}{2}c\)     AG

[4 marks]

a.

(i)     \(\overrightarrow {{\text{RA}}}  = \frac{1}{3}\overrightarrow {{\text{BA}}} \)

\( = \frac{1}{3}\)(ab)     A1

(ii)     \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \)

\( = \frac{2}{3}\left( {\overrightarrow {{\text{RA}}}  + \overrightarrow {{\text{AS}}} } \right)\)     (M1)

\( = \frac{2}{3}\left( {\frac{1}{3}(a – b) + \frac{2}{3}(c – a)} \right)\;\;\;\)or equivalent.     A1A1

\( = \frac{2}{9}\)\(\left( {a – b} \right)\) \( + \frac{4}{9}\)\(\left( {c – a} \right)\)     A1

\(\overrightarrow {{\text{RT}}}  =  – \frac{2}{9}\)\(a\) – \( – \frac{2}{9}\)\(b\) \( + \frac{4}{9}\)\(c\)     AG

[5 marks]

b.

\(\overrightarrow {{\text{BT}}}  = \overrightarrow {{\text{BR}}}  + \overrightarrow {{\text{RT}}} \)

\( = \frac{2}{3}\overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{RT}}} \)     (M1)

\( = \frac{2}{3}a – \frac{2}{3}b – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\)    A1

\(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\left( {\frac{1}{2}a – b + \frac{1}{2}c} \right)\)     A1

point \(B\) is common to \(\overrightarrow {{\text{BT}}} \) and \(\overrightarrow {{\text{BM}}} \) and \(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\overrightarrow {{\text{BM}}} \)     R1R1

so \(T\) lies on [\(BM\)]     AG

[5 marks]

Total [14 marks]

c.

Examiners report

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

a.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

b.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

c.

Question

The position vectors of the points \(A\), \(B\) and \(C\) are \(a\), \(b\) and \(c\) respectively, relative to an origin \(O\). The following diagram shows the triangle \(ABC\) and points \(M\), \(R\), \(S\) and \(T\).

\(M\) is the midpoint of [\(AC\)].

\(R\) is a point on [\(AB\)] such that \(\overrightarrow {{\text{AR}}}  = \frac{1}{3}\overrightarrow {{\text{AB}}} \).

\(S\) is a point on [\(AC\)] such that \(\overrightarrow {{\text{AS}}}  = \frac{2}{3}\overrightarrow {{\text{AC}}} \).

\(T\) is a point on [\(RS\)] such that \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \).

(i)     Express \(\overrightarrow {{\text{AM}}} \) in terms of \(a\) and \(c\).

(ii)     Hence show that \(\overrightarrow {{\text{BM}}}  = \frac{1}{2}\)\(a\) – \(b\)\( + \frac{1}{2}c\).

[4]
a.

(i)     Express \(\overrightarrow {{\text{RA}}} \) in terms of \(a\) and \(b\).

(ii)     Show that \(\overrightarrow {RT}  =  – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\).

[5]
b.

Prove that \(T\) lies on [\(BM\)].

[5]
c.
Answer/Explanation

Markscheme

(i)     \(\overrightarrow {{\text{AM}}}  = \frac{1}{2}\overrightarrow {{\text{AC}}} \)     (M1)

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

(ii)     \(\overrightarrow {{\text{BM}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AM}}} \)     M1

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

\(\overrightarrow {{\text{BM}}}  = \frac{1}{2}a – b + \frac{1}{2}c\)     AG

[4 marks]

a.

(i)     \(\overrightarrow {{\text{RA}}}  = \frac{1}{3}\overrightarrow {{\text{BA}}} \)

\( = \frac{1}{3}\)(ab)     A1

(ii)     \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \)

\( = \frac{2}{3}\left( {\overrightarrow {{\text{RA}}}  + \overrightarrow {{\text{AS}}} } \right)\)     (M1)

\( = \frac{2}{3}\left( {\frac{1}{3}(a – b) + \frac{2}{3}(c – a)} \right)\;\;\;\)or equivalent.     A1A1

\( = \frac{2}{9}\)\(\left( {a – b} \right)\) \( + \frac{4}{9}\)\(\left( {c – a} \right)\)     A1

\(\overrightarrow {{\text{RT}}}  =  – \frac{2}{9}\)\(a\) – \( – \frac{2}{9}\)\(b\) \( + \frac{4}{9}\)\(c\)     AG

[5 marks]

b.

\(\overrightarrow {{\text{BT}}}  = \overrightarrow {{\text{BR}}}  + \overrightarrow {{\text{RT}}} \)

\( = \frac{2}{3}\overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{RT}}} \)     (M1)

\( = \frac{2}{3}a – \frac{2}{3}b – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\)    A1

\(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\left( {\frac{1}{2}a – b + \frac{1}{2}c} \right)\)     A1

point \(B\) is common to \(\overrightarrow {{\text{BT}}} \) and \(\overrightarrow {{\text{BM}}} \) and \(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\overrightarrow {{\text{BM}}} \)     R1R1

so \(T\) lies on [\(BM\)]     AG

[5 marks]

Total [14 marks]

c.

Examiners report

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

a.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

b.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

c.

Question

The position vectors of the points \(A\), \(B\) and \(C\) are \(a\), \(b\) and \(c\) respectively, relative to an origin \(O\). The following diagram shows the triangle \(ABC\) and points \(M\), \(R\), \(S\) and \(T\).

\(M\) is the midpoint of [\(AC\)].

\(R\) is a point on [\(AB\)] such that \(\overrightarrow {{\text{AR}}}  = \frac{1}{3}\overrightarrow {{\text{AB}}} \).

\(S\) is a point on [\(AC\)] such that \(\overrightarrow {{\text{AS}}}  = \frac{2}{3}\overrightarrow {{\text{AC}}} \).

\(T\) is a point on [\(RS\)] such that \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \).

(i)     Express \(\overrightarrow {{\text{AM}}} \) in terms of \(a\) and \(c\).

(ii)     Hence show that \(\overrightarrow {{\text{BM}}}  = \frac{1}{2}\)\(a\) – \(b\)\( + \frac{1}{2}c\).

[4]
a.

(i)     Express \(\overrightarrow {{\text{RA}}} \) in terms of \(a\) and \(b\).

(ii)     Show that \(\overrightarrow {RT}  =  – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\).

[5]
b.

Prove that \(T\) lies on [\(BM\)].

[5]
c.
Answer/Explanation

Markscheme

(i)     \(\overrightarrow {{\text{AM}}}  = \frac{1}{2}\overrightarrow {{\text{AC}}} \)     (M1)

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

(ii)     \(\overrightarrow {{\text{BM}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AM}}} \)     M1

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

\(\overrightarrow {{\text{BM}}}  = \frac{1}{2}a – b + \frac{1}{2}c\)     AG

[4 marks]

a.

(i)     \(\overrightarrow {{\text{RA}}}  = \frac{1}{3}\overrightarrow {{\text{BA}}} \)

\( = \frac{1}{3}\)(ab)     A1

(ii)     \(\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}} \)

\( = \frac{2}{3}\left( {\overrightarrow {{\text{RA}}}  + \overrightarrow {{\text{AS}}} } \right)\)     (M1)

\( = \frac{2}{3}\left( {\frac{1}{3}(a – b) + \frac{2}{3}(c – a)} \right)\;\;\;\)or equivalent.     A1A1

\( = \frac{2}{9}\)\(\left( {a – b} \right)\) \( + \frac{4}{9}\)\(\left( {c – a} \right)\)     A1

\(\overrightarrow {{\text{RT}}}  =  – \frac{2}{9}\)\(a\) – \( – \frac{2}{9}\)\(b\) \( + \frac{4}{9}\)\(c\)     AG

[5 marks]

b.

\(\overrightarrow {{\text{BT}}}  = \overrightarrow {{\text{BR}}}  + \overrightarrow {{\text{RT}}} \)

\( = \frac{2}{3}\overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{RT}}} \)     (M1)

\( = \frac{2}{3}a – \frac{2}{3}b – \frac{2}{9}a – \frac{2}{9}b + \frac{4}{9}c\)    A1

\(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\left( {\frac{1}{2}a – b + \frac{1}{2}c} \right)\)     A1

point \(B\) is common to \(\overrightarrow {{\text{BT}}} \) and \(\overrightarrow {{\text{BM}}} \) and \(\overrightarrow {{\text{BT}}}  = \frac{8}{9}\overrightarrow {{\text{BM}}} \)     R1R1

so \(T\) lies on [\(BM\)]     AG

[5 marks]

Total [14 marks]

c.

Examiners report

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

a.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

b.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

c.

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 \(\overrightarrow {{\text{OC}}}  = {{c}}\).

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

[2]
a.

(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.

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]
d.
Answer/Explanation

Markscheme

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

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

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

[2 marks]

a.

(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]

b.

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

c.

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

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

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 \(\lambda ,{\text{ }}\mu  \in \mathbb{R}\).

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

[1]
a.i.

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

[2]
a.ii.

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

[2]
b.i.

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

[2]
b.ii.

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

[4]
c.

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

[2]
d.

Given that area \(\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})\), find the value of \(k\).

[5]
e.
Answer/Explanation

Markscheme

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

[1 mark]

a.i.

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

\( = \frac{1}{7}\)ba     A1

[2 marks]

a.ii.

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

[2 marks]

b.i.

\(\overrightarrow {{\text{OD}}}  = \frac{1}{2}\) 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]

b.ii.

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]

c.

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

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

[2 marks]

d.

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]

e.

Examiners report

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a.i.

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a.ii.

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b.i.

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b.ii.

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c.

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d.

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e.

Question: [Maximum mark: 9]

Consider the vectors a and b such that \(a = \binom{12}{-5} and |b| = 15\)
(a) Find the possible range of values for | a + b |.
Consider the vector p such that p = a + b.
(b) Given that | a + b | is a minimum, find p.
Consider the vector q such that \(q =\binom{x}{y}\), where x , y ∈ R+ .
(c) Find q such that |q| = |b| and q is perpendicular to a.

Answer/Explanation

Ans:

Note: Award (A1)A0 for 2 and 28 seen with no indication that they are the endpoints of an interval.

(b) recognition that p or b is a negative multiple of a 

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