# IBDP Maths AA: Topic : AHL 3.12: sum and difference of two vectors: IB style Questions HL Paper 2

### Question

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.

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

### Question

The following diagram shows two perpendicular vectors u and v.

a.Let $$w = u – v$$. Represent $$w$$ on the diagram above.[2]

b.Given that $$u = \left( \begin{array}{c}3\\2\\1\end{array} \right)$$ and $$v = \left( \begin{array}{c}5\\n\\3\end{array} \right)$$, where $$n \in \mathbb{Z}$$, find $$n$$.[4]

## Markscheme

METHOD 1 A1A1      N2

Note: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).

METHOD 2

A1A1      N2

Notes: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).

[2 marks]

a.

evidence of setting scalar product equal to zero (seen anywhere)     R1

eg   u $$\cdot$$ v $$= 0,{\text{ }}15 + 2n + 3 = 0$$

correct expression for scalar product     (A1)

eg   $$3 \times 5 + 2 \times n + 1 \times 3,{\text{ }}2n + 18 = 0$$

attempt to solve equation     (M1)

eg   $$2n = – 18$$

$$n = – 9$$     A1     N3

[4 marks]

## Question

Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.

Let $${\mathop {{\text{PR}}}\limits^ \to }$$ = 6i − j + 3k.

a.i.Find $$\mathop {{\text{PQ}}}\limits^ \to$$.[2]

a.ii.Find $$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|$$.[2]

b.Find the angle between PQ and PR.[4]

c.Find the area of triangle PQR.[2

d.Hence or otherwise find the shortest distance from R to the line through P and Q.[3]

## Markscheme

valid approach      (M1)

eg   (7, 4, 9) − (3, 2, 5)  A − B

$$\mathop {{\text{PQ}}}\limits^ \to =$$ 4i + 2j + 4k $$\left( { = \left( \begin{gathered} 4 \hfill \\ 2 \hfill \\ 4 \hfill \\ \end{gathered} \right)} \right)$$     A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula      (A1)
eg  $$\sqrt {{4^2} + {2^2} + {4^2}}$$

$$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6$$     A1 N2

[2 marks]

a.ii.

finding scalar product and magnitudes      (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR = $$\sqrt {36 + 1 + 9} = \left( {6.782} \right)$$

correct substitution of their values to find cos $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$     M1

eg  cos $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 – 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355$$

0.581746

$${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$ = 0.582 radians  or $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$ = 33.3°     A1 N3

[4 marks]

b.

correct substitution (A1)
eg    $$\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582$$

area is 11.2 (sq. units)      A1 N2

[2 marks]

c.

recognizing shortest distance is perpendicular distance from R to line through P and Q      (M1)

eg  sketch, height of triangle with base $$\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}$$

correct working      (A1)

eg  $$\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ$$

3.72677

distance = 3.73  (units)    A1 N2

[3 marks]

## Question

Let  $${\boldsymbol{v}} = 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}$$ and $${\boldsymbol{w}} = {\boldsymbol{i}} + 2{\boldsymbol{j}} – 3{\boldsymbol{k}}$$ . The vector $${\boldsymbol{v}} + p{\boldsymbol{w}}$$ is perpendicular to w. Find the value of p.

## Markscheme

$$p{\boldsymbol{w}} = p{\boldsymbol{i}} + 2p{\boldsymbol{j}} – 3p{\boldsymbol{k}}$$ (seen anywhere)     (A1)

attempt to find $${\boldsymbol{v}} + p{\boldsymbol{w}}$$     (M1)

e.g. $$3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} + p({\boldsymbol{i}} + 2{\boldsymbol{j}} – 3{\boldsymbol{k}})$$

collecting terms $$(3 + p){\boldsymbol{i}} + (4 + 2p){\boldsymbol{j}} + (1 – 3p){\boldsymbol{k}}$$     A1

attempt to find the dot product     (M1)

e.g. $$1(3 + p) + 2(4 + 2p) – 3(1 – 3p)$$

setting their dot product equal to 0     (M1)

e.g. $$1(3 + p) + 2(4 + 2p) – 3(1 – 3p) = 0$$

simplifying     A1

e.g. $$3 + p + 8 + 4p – 3 + 9p = 0$$ , $$14p + 8 = 0$$

$$p = – 0.571$$ $$\left( { – \frac{8}{{14}}} \right)$$     A1     N3

[7 marks]

## Question

Let $$u = 6i + 3j + 6k$$ and $$v = 2i + 2j + k$$.

a.Find

(i)     $$u \bullet v$$;

(ii)     $$\left| {{u}} \right|$$;

(iii)     $$\left| {{v}} \right|$$.[5]

b.Find the angle between $${{u}}$$ and $${{v}}$$.[2]

## Markscheme

(i)     correct substitution     (A1)

eg$$\;\;\;6 \times 2 + 3 \times 2 + 6 \times 1$$

$$u \bullet v = 24$$     A1     N2

(ii)     correct substitution into magnitude formula for $${{u}}$$ or $${{v}}$$     (A1)

eg$$\;\;\;\sqrt {{6^2} + {3^2} + {6^2}} ,{\text{ }}\sqrt {{2^2} + {2^2} + {1^2}}$$, correct value for $$\left| {{v}} \right|$$

$$\left| {{u}} \right| = 9$$     A1     N2

(iii)     $$\left| {{v}} \right| = 3$$     A1     N1

[5 marks]

a.

correct substitution into angle formula     (A1)

eg$$\;\;\;\frac{{24}}{{9 \times 3}},{\text{ }}0.\bar 8$$

$$0.475882,{\text{ }}27.26604^\circ$$     A1     N2

$$0.476,{\text{ }}27.3^\circ$$

[2 marks]

Total [7 marks]

b.

## Question

The diagram shows a parallelogram ABCD.

The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .

(i)     Show that $$\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right)$$ .

(ii)    Find $$\overrightarrow {{\rm{AD}}}$$ .

(iii)   Hence show that $$\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 6\\ 5\\ 3 \end{array}} \right)$$ .
[5]

a(i), (ii) and (iii).

Find the coordinates of point C.[3]

c(i) and (ii).

(i)     Find $$\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}}$$.

(ii)    Hence find angle A.[7]

d

Hence, or otherwise, find the area of the parallelogram.[3]

## Markscheme

(i) evidence of approach     M1

e.g. $${\text{B}} – {\text{A}}$$ , $$\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}}$$ , $$\left( {\begin{array}{*{20}{c}} 6\\ 4\\ 4 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right)$$

$$\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right)$$     AG     N0

(ii) evidence of approach     (M1)

e.g. $${\text{D}} – {\text{A}}$$ , $$\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}}$$ , $$\left( {\begin{array}{*{20}{c}} 2\\ 5\\ 5 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right)$$

$$\overrightarrow {{\rm{AD}}}= \left( {\begin{array}{*{20}{c}} 1\\ 3\\ 2 \end{array}} \right)$$     A1     N2

(iii) evidence of approach     (M1)

e.g. $$\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{AD}}}$$

correct substitution     A1

e.g. $$\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1\\ 3\\ 2 \end{array}} \right)$$

$$\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 6\\ 5\\ 3 \end{array}} \right)$$     AG     N0

[5 marks]

a(i), (ii) and (iii).

evidence of combining vectors (there are at least 5 ways)     (M1)

e.g. $$\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AC}}}$$ , $$\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{AD}}}$$,  $$\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OD}}}$$

correct substitution     A1

$$\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6\\ 5\\ 3 \end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}} 7\\ 7\\ 6 \end{array}} \right)} \right)$$

e.g. coordinates of C are $$(7{\text{, }}7{\text{, }}6)$$     A1     N1

[3 marks]

b.

(i) evidence of using scalar product on $$\overrightarrow {{\rm{AB}}}$$ and $$\overrightarrow {{\rm{AD}}}$$    (M1)

e.g. $$\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 5(1) + 2(3) + 1(2)$$

$$\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 13$$    A1     N2

(ii) $$\left| {\overrightarrow {{\rm{AB}}} } \right| = 5.477 \ldots$$ , $$\left| {\overrightarrow {{\rm{AD}}} } \right| = 3.741 \ldots$$    (A1)(A1)

evidence of using $$\cos A = \frac{{\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} }}{{\left| {\overrightarrow {{\rm{AB}}} } \right|\left| {\overrightarrow {{\rm{AD}}} } \right|}}$$     (M1)

correct substitution     A1

e.g. $$\cos A = \frac{{13}}{{20.493}}$$

$$\widehat A = 0.884$$ $$(50.6^\circ )$$     A1     N3

[7 marks]

c(i) and (ii).

METHOD 1

evidence of using $${\rm{area}} = 2\left( {\frac{1}{2}\left| {\overrightarrow {{\rm{AD}}} } \right|\left| {\overrightarrow {{\rm{AB}}} } \right|\sin {\rm{D}}\widehat {\rm{A}}{\rm{B}}} \right)$$     (M1)

correct substitution     A1

e.g. $${\rm{area}} = 2\left( {\frac{1}{2}(3,741 \ldots )(5.477 \ldots )\sin 0.883 \ldots } \right)$$

$${\rm{area}} = 15.8$$     A1     N2

METHOD 2

evidence of using $${\rm{area}} = b \times h$$     (M1)

finding height of parallelogram     A1

e.g. $$h = 3.741 \ldots \times \sin 0.883 \ldots ( = 2.892 \ldots )$$ , $$h = 5.477 \ldots \times \sin 0.883 \ldots ( = 4.234 \ldots )$$

$${\rm{area}} = 15.8$$     A1     N2

[3 marks]

## Question

Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let $${L_1}$$ be the line through P and Q.

a.Show that $$\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}} 1\\ { – 2}\\ 3 \end{array}} \right)$$ .
[1]

b.

The line $${L_1}$$ may be represented by $${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 3 \\ { – 3} \\ 8 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1 \\ { – 2} \\ 3 \end{array}} \right)$$ .

(i)     What information does the vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 3}\\ 8 \end{array}} \right)$$ give about $${L_1}$$ ?

(ii)    Write down another vector representation for $${L_1}$$ using $$\left( {\begin{array}{*{20}{c}} 3\\ { – 3}\\ 8 \end{array}} \right)$$ .
[3]

c.

The point $${\text{T}}( – 1{\text{, }}5{\text{, }}p)$$ lies on $${L_1}$$ .

Find the value of $$p$$ .[3]

d.

The point T also lies on $${L_2}$$ with equation $$\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 3}\\ 9\\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { – 2}\\ q \end{array}} \right)$$ .

Show that $$q = – 3$$ .

[3]
e.

Let $$\theta$$ be the obtuse angle between $${L_1}$$ and $${L_2}$$ . Calculate the size of $$\theta$$ .[7]

## Markscheme

evidence of correct approach     A1

e.g. $$\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} – \overrightarrow {{\rm{OP}}}$$ , $$\left( {\begin{array}{*{20}{c}} 3\\ { – 3}\\ 8 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 2\\ { – 1}\\ 5 \end{array}} \right)$$

$$\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}} 1\\ { – 2}\\ 3 \end{array}} \right)$$     AG     N0

[1 mark]

a.

(i) correct description     R1     N1

e.g. reference to $$\left( {\begin{array}{*{20}{c}} 3\\ { – 3}\\ 8 \end{array}} \right)$$ being the position vector of a point on the line, a vector to the line, a point on the line.

(ii) any correct expression in the form $${\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}$$    A2     N2

where $${\boldsymbol{a}}$$ is $$\left( {\begin{array}{*{20}{c}} 3 \\ { – 3} \\ 8 \end{array}} \right)$$ , and $${\boldsymbol{b}}$$ is a scalar multiple of $$\left( {\begin{array}{*{20}{c}} 1 \\ { – 2} \\ 3 \end{array}} \right)$$

e.g. $${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 3 \\ { – 3} \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ { – 3} \end{array}} \right)$$ , $${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} {3 + 2s} \\ { – 3 – 4s} \\ {8 + 6s} \end{array}} \right)$$

[3 marks]

b.

one correct equation     (A1)

e.g. $$3 + s = – 1$$ , $$– 3 – 2s = 5$$

$$s = – 4$$     A1

$$p = – 4$$     A1     N2

[3 marks]

c.

one correct equation     A1

e.g. $$– 3 + t = – 1$$ , $$9 – 2t = 5$$

$$t = 2$$     A1

substituting $$t = 2$$

e.g. $$2 + 2q = – 4$$ , $$2q = – 6$$     A1

$$q = – 3$$     AG     N0

[3 marks]

d.

choosing correct direction vectors $$\left( {\begin{array}{*{20}{c}} 1\\ { – 2}\\ 3 \end{array}} \right)$$ and $$\left( {\begin{array}{*{20}{c}} 1\\ { – 2}\\ { – 3} \end{array}} \right)$$     (A1)(A1)

finding correct scalar product and magnitudes     (A1)(A1)(A1)

scalar product $$(1)(1) + ( – 2)( – 2) + ( – 3)(3)$$     $$( = – 4)$$

magnitudes $$\sqrt {{1^2} + {{( – 2)}^2} + {3^2}}$$ $$= \sqrt {14}$$ , $$\sqrt {{1^2} + {{( – 2)}^2} + {{( – 3)}^2}}$$ $$= \sqrt {14}$$

evidence of substituting into scalar product     M1

e.g. $$\cos \theta = \frac{{ – 4}}{{3.741 \ldots \times 3.741 \ldots }}$$

$$\theta = 1.86$$ radians (or $$107^\circ$$)    A1     N4

[7 marks]

## Question

Consider the lines $${L_1}$$ and $${L_2}$$ with equations $${L_1}$$ : $$\boldsymbol{r}=\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right)$$ and $${L_2}$$ : $$\boldsymbol{r} = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)$$.

The lines intersect at point $$\rm{P}$$.

a.Find the coordinates of $${\text{P}}$$.[6]

b.Show that the lines are perpendicular.[5]

c.The point $${\text{Q}}(7, 5, 3)$$ lies on $${L_1}$$. The point $${\text{R}}$$ is the reflection of $${\text{Q}}$$ in the line $${L_2}$$.

Find the coordinates of $${\text{R}}$$.[6]

## Markscheme

appropriate approach     (M1)

eg     $$\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)$$, $${L_1} = {L_2}$$

any two correct equations     A1A1

eg     $$11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 – s = – 7 + 11t$$

attempt to solve system of equations     (M1)

eg     $$10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s – 2t = – 10} \\ {3s – t = – 7} \end{array}} \right.$$

one correct parameter     A1

eg     $$s = – 2,{\text{ }}t = 1$$

$${\text{P}}(3, 2, 4)$$   (accept position vector)     A1     N3

[6 marks]

a.

choosing correct direction vectors for $${L_1}$$ and $${L_2}$$     (A1)(A1)

eg     $$\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ {11} \end{array}} \right)$$ (or any scalar multiple)

evidence of scalar product (with any vectors)     (M1)

eg     $$a \cdot b$$, $$\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)$$

correct substitution     A1

eg     $$4(2) + 3(1) + ( – 1)(11),{\text{ }}8 + 3 – 11$$

calculating $$a \cdot b = 0$$     A1

Note: Do not award the final A1 without evidence of calculation.

vectors are perpendicular     AG     N0

[5 marks]

b.

Note: Candidates may take different approaches, which do not necessarily involve vectors.

In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.

METHOD 1

attempt to find $$\overrightarrow {{\text{QP}}}$$ or $$\overrightarrow {{\text{PQ}}}$$     (M1)

correct working (may be seen on diagram)     A1

eg     $$\overrightarrow {{\text{QP}}}$$ = $$\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right)$$, $$\overrightarrow {{\text{PQ}}}$$ = $$\left( \begin{array}{c}7\\5\\3\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right)$$

recognizing $${\text{R}}$$ is on $${L_1}$$ (seen anywhere)     (R1)

eg     on diagram

$${\text{Q}}$$ and $${\text{R}}$$ are equidistant from $${\text{P}}$$ (seen anywhere)     (R1)

eg     $$\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{PR}}}$$, marked on diagram

correct working     (A1)

eg     $$\left( \begin{array}{c}3\\2\\4\end{array} \right) – \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)$$

$${\text{R}}(–1, –1, 5)$$ (accept position vector)     A1     N3

METHOD 2

recognizing $${\text{R}}$$ is on $${L_1}$$ (seen anywhere)     (R1)

eg     on diagram

$${\text{Q}}$$ and $${\text{R}}$$ are equidistant from $${\text{P}}$$ (seen anywhere)     (R1)

eg     $${\text{P}}$$ midpoint of $${\text{QR}}$$, marked on diagram

valid approach to find one coordinate of mid-point     (M1)

eg     $${x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)$$

one correct substitution     A1

eg     $${x_R} = 3 + (3 – 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)$$

correct working for one coordinate     (A1)

eg     $${x_R} = 3 – 4,{\text{ }}4 – 5 = {y_R},{\text{ }}8 = (z + 3)$$

$${\text{R}} (-1, -1, 5)$$ (accept position vector)     A1     N3

[6 marks]

c.
c.

## Question

Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.

Let $${\mathop {{\text{PR}}}\limits^ \to }$$ = 6i − j + 3k.

a.i.Find $$\mathop {{\text{PQ}}}\limits^ \to$$.[2]

a.ii.Find $$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|$$.[2]

b.Find the angle between PQ and PR.[4]

c.Find the area of triangle PQR.[2]

d.Hence or otherwise find the shortest distance from R to the line through P and Q.[3]

## Markscheme

valid approach (M1)

eg (7, 4, 9) − (3, 2, 5) A − B

$$\mathop {{\text{PQ}}}\limits^ \to =$$ 4i + 2j + 4k $$\left( { = \left( \begin{gathered} 4 \hfill \\ 2 \hfill \\ 4 \hfill \\ \end{gathered} \right)} \right)$$ A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula (A1)
eg $$\sqrt {{4^2} + {2^2} + {4^2}}$$

$$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6$$ A1 N2

[2 marks]

a.ii.

finding scalar product and magnitudes (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR = $$\sqrt {36 + 1 + 9} = \left( {6.782} \right)$$

correct substitution of their values to find cos $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$ M1

eg cos $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 – 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355$$

0.581746

$${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$ = 0.582 radians or $${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$$ = 33.3° A1 N3

[4 marks]

b.

correct substitution (A1)
eg $$\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582$$

area is 11.2 (sq. units) A1 N2

[2 marks]

c.

recognizing shortest distance is perpendicular distance from R to line through P and Q (M1)

eg sketch, height of triangle with base $$\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}$$

correct working (A1)

eg $$\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ$$

3.72677

distance = 3.73 (units) A1 N2

[3 marks]

d.

## Question

Consider the points A($$5$$, $$2$$, $$1$$) , B($$6$$, $$5$$, $$3$$) , and C($$7$$, $$6$$, $$a + 1$$) , $$a \in{\mathbb{R}}$$ .

Let $${\rm{q}}$$ be the angle between $$\overrightarrow {{\rm{AB}}}$$ and $$\overrightarrow {{\rm{AC}}}$$ .

a.Find

(i)     $$\overrightarrow {{\rm{AB}}}$$ ;

(ii)     $$\overrightarrow {{\rm{AC}}}$$ .[3]

b.Find the value of $$a$$ for which $${\rm{q}} = \frac{\pi }{2}$$ .[4]

c.i. Show that $$\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}$$ .

ii. Hence, find the value of a for which $${\rm{q}} = 1.2$$ .[8]

c.ii.Hence, find the value of a for which $${\rm{q}} = 1.2$$ .[4]

(i)     appropriate approach     (M1)

eg   $$\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}}$$ ,  $${\rm{B}} – {\rm{A}}$$

$$\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l} 1\\ 3\\ 2 \end{array} \right)$$     A1     N2

(ii)     $$\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l} 2\\ 4\\ a \end{array} \right)$$     A1     N1

[3 marks]

a.

valid reasoning (seen anywhere)     R1

eg   scalar product is zero, $$\cos \frac{\pi }{2} = \frac{{\boldsymbol{u} \bullet \boldsymbol{v}}}{{\left| \boldsymbol{u} \right|\left| \boldsymbol{v} \right|}}$$

correct scalar product of their $$\overrightarrow {{\rm{AB}}}$$ and $$\overrightarrow {{\rm{AC}}}$$ (may be seen in part (c))     (A1)

eg   $$1(2) + 3(4) + 2(a)$$

correct working for their $$\overrightarrow {{\rm{AB}}}$$ and $$\overrightarrow {{\rm{AC}}}$$     (A1)

eg   $$2a + 14$$ , $$2a = – 14$$

$$a = – 7$$    A1     N3

[4 marks]

b.

correct magnitudes (may be seen in (b))     (A1)(A1)

$$\sqrt {{1^2} + {3^2} + {2^2}} \left( { = \sqrt {14} } \right)$$ , $$\sqrt {{2^2} + {4^2} + {a^2}} \left( { = \sqrt {20 + {a^2}} } \right)$$

substitution into formula     (M1)

eg   $$\cos \theta \frac{{1 \times 2 + 3 \times 4 + 2 \times a}}{{\sqrt {{1^2} + {3^2} + {2^2}} \sqrt {{2^2} + {4^2} + {a^2}} }}$$ , $$\frac{{14 + 2a}}{{\sqrt {14} \sqrt {4 + 16 + {a^2}} }}$$

eg   $$\cos \theta = \frac{{14 + 2a}}{{\sqrt {14} \sqrt {20 + {a^2}} }}$$

$$\cos \theta = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}$$     AG     N0

[4 marks]

correct setup     (A1)

eg   $$\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}$$

valid attempt to solve     (M1)

eg sketch, $$\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0$$ , attempt to square

$$a = – 3.25$$     A2     N3

[4 marks]

c.

correct setup     (A1)

eg   $$\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}$$

valid attempt to solve     (M1)

eg sketch, $$\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0$$ , attempt to square

$$a = – 3.25$$     A2     N3

[4 marks]

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