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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
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 \(\mathop {{\text{PQ}}}\limits^ \to \).
Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|\).
Find the angle between PQ and PR.
Find the area of triangle PQR.
Hence or otherwise find the shortest distance from R to the line through P and Q.
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]
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]
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]
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]
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]