IB DP Maths Topic 4.1 Algebraic and geometric approaches to the sum and difference of two vectors SL Paper 2

 

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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

The following diagram shows two perpendicular vectors u and v.

Let \(w = u – v\). Represent \(w\) on the diagram above.

[2]
a.

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]
b.

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]

b.

Examiners report

[N/A]
a.
[N/A]
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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