Find

(i)     \(\overrightarrow {{\rm{PQ}}} \) ;

(ii)    \(\overrightarrow {{\rm{PR}}} \) .

Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .

(i)     Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .

(ii)    Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .

(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]

Combining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated the magnitudes.

Many candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used the cosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks. Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established.

Many candidates attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew the side relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of \(1,2,\sqrt 3 \) to find the area of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected as the value of sine was expected to be used in the approach.