IB DP Maths Topic 4.1 Algebraic and geometric approaches to unit vectors; base vectors; i, j and k. SL Paper 2

 

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Question

The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .

(i)     Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .

(ii)    Find \({\rm{B}}\widehat {\rm{A}}{\rm{O}}\) .

[8]
a(i) and (ii).

The line \({L_1}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .

Write down the coordinates of two points on \({L_1}\) .

[2]
b.

The line \({L_2}\) passes through A and is parallel to \(\overrightarrow {{\rm{OB}}} \) .

(i)     Find a vector equation for \({L_2}\) , giving your answer in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) . 

(ii)    Point \(C(k, – k,5)\) is on  \({L_2}\) . Find the coordinates of C.

[6]
c(i) and (ii).

The line \({L_3}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) and passes through the point C. 

 Find the value of p at C.

[2]
d.
Answer/Explanation

Markscheme

(i) evidence of approach     M1

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}} \)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\)    AG     N0

(ii) for choosing correct vectors, (\(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{OA}}} \) with \(\overrightarrow {{\rm{BA}}} \) )     (A1)(A1)

Note: Using \(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{BA}}} \) will lead to \(\pi  – 0.799\) . If they then say \({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) , this is a correct solution.

calculating \(\overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} \) , \(\left| {\overrightarrow {{\rm{AO}}} } \right|\) , \(\left| {\overrightarrow {{\rm{AB}}} } \right|\)     (A1)(A1)(A1)

e.g. \({d_1} \bullet {d_2} = ( – 1)( – 4) + (2)(6) + ( – 3)( – 1)( = 19)\)

\(\left| {{d_1}} \right| = \sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} ( = \sqrt {14} )\) , \(\left| {{d_2}} \right| = \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} ( = \sqrt {53} )\)

evidence of using the formula to find the angle     M1

e.g. \(\cos \theta = \frac{{( – 1)( – 4) + (2)(6) + ( – 3)( – 1)}}{{\sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} }}\) , \(\frac{{19}}{{\sqrt {14} \sqrt {53} }}\) , \(0.69751 \ldots \)

\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) radians (accept \(45.8^\circ \))     A1     N3

[8 marks]

a(i) and (ii).

two correct answers     A1A1

e.g. (1, \( – 2\), 3) , (\( – 3\), 4, 2) , (\( – 7\), 10, 1), (\( – 11\), 16, 0)     N2

[2 marks]

b.

(i) \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\)     A2     N2

(ii) C on \({L_2}\) , so \(\left( {\begin{array}{*{20}{c}}
k\\
{ – k}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\)     (M1)

evidence of equating components (A1)

e.g. \(1 – 3t = k\) , \( – 2 + 4t = – k\) , \(5 = 3 + 2t\)

one correct value \(t = 1\) , \(k = – 2\) (seen anywhere)     (A1)

coordinates of C are \(( – 2{\text{, }}2{\text{, }}5)\)     A1     N3

[6 marks]

c(i) and (ii).

for setting up one (or more) correct equation using \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\)     (M1)

e.g. \(3 + p = – 2\) , \( – 8 – 2p = 2\) , \( – p = 5\)

\(p = – 5\)     A1     N2

[2 marks]

d.

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]
b.

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

(ii)    Hence find angle A.

[7]
c(i) and (ii).

Hence, or otherwise, find the area of the parallelogram.

[3]
d.
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]

d.

Question

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

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

[1]
a.

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]
b.

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

Find the value of \(p\) .

[3]
c.

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]
d.

Let \(\theta \) be the obtuse angle between \({L_1}\) and \({L_2}\) . Calculate the size of \(\theta \) .

[7]
e.
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]

e.
e.

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 \(\rm{P}\).

Find the coordinates of \({\text{P}}\).

[6]
a.

Show that the lines are perpendicular.

[5]
b.

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]
c.
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

Let \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)\).

Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).

[2]
a.

Let \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)\). Find \({\rm{B\hat AC}}\).

[4]
b.
Answer/Explanation

Markscheme

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\sqrt {{4^2} + {1^2} + {2^2}} \)

4.58257

\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21} \) (exact), 4.58     A1     N2

[2 marks]

a.

finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\)     (A1)(A1)

scalar product \( = (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)\)

\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + 0 + 0} {\text{ }}( = 3)\)

substituting their values into cosine formula     (M1)

eg cos B\(\hat A\)C\({\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}}  \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta  = 0.873\)

0.509739 (29.2059°)

\({\rm{B\hat AC}} = 0.510\) (29.2°)     A1     N2

[4 marks]

b.
b.

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.

Find \(\mathop {{\text{PQ}}}\limits^ \to \).

[2]
a.i.

Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|\).

[2]
a.ii.

Find the angle between PQ and PR.

[4]
b.

Find the area of triangle PQR.

[2]
c.

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

[3]
d.
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.
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