IB DP Maths Topic 4.1 Algebraic and geometric approaches to the sum and difference of two vectors SL Paper 1

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{PO}}}  + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} – {\rm{P}}\)

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

(ii) \(\overrightarrow {{\rm{PR}}}  = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\)     A1     N1

[3 marks]

METHOD 1

choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \)    (A1)(A1)

finding \(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\)    (A1) (A1)(A1)

\(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}}  = – 2 + 4 + 4( = 6)\)

\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( – 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)

substituting into formula for angle between two vectors     M1

e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6  \times \sqrt {24} }}\)

simplifying to expression clearly leading to \(\frac{1}{2}\)     A1

e.g. \(\frac{6}{{\sqrt 6  \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)     AG     N0

METHOD 2

evidence of choosing cosine rule (seen anywhere)     (M1)

\(\overrightarrow {{\rm{QR}}}  = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\)     A1

\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \)     (A1)(A1)(A1)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} – {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6  \times \sqrt {24} }}\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 – 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)      AG     N0

[7 marks]

(i) METHOD 1

evidence of appropriate approach     (M1)

e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram

substituting correctly     (A1)

e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} \)

\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\)     A1     N3

METHOD 2

since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \)     (A1)

evidence of approach

e.g. drawing a right triangle, finding the missing side     (A1)

\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\)     A1     N3

(ii) evidence of appropriate approach      (M1)

e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)

correct substitution

e.g. area \( = \frac{1}{2}\sqrt 6  \times \sqrt {24}  \times \frac{{\sqrt 3 }}{2}\)     A1

area \( = 3\sqrt 3 \)     A1     N2

[6 marks]