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.

▶️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 

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]

 
▶️Answer/Explanation

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).

Additional lines not required.

[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]

▶️Answer/Explanation

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.

Answer/Explanation

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]

 
▶️Answer/Explanation

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]

▶️Answer/Explanation

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]

▶️Answer/Explanation

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]

 
▶️Answer/Explanation

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]

▶️Answer/Explanation

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]

▶️Answer/Explanation

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

simplification leading to required answer     A1

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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