Question
OACB is a parallelogram with \(\overrightarrow {{\text{OA}}} = \) a and \(\overrightarrow {{\text{OB}}} = \) b, where a and b are non-zero vectors.
Show that
(i) \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);
(ii) \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.
Answer/Explanation
Markscheme
METHOD 1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}} \)
= (a + b) \( \bullet \) (a + b) A1
= a \( \bullet \) a + a \( \bullet \) b + b \( \bullet \) a + b \( \bullet \) b A1
= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\) A1
\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = – \)(a \( \bullet \) b) A1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
(ii) METHOD 1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}} \)
= (b − a) \( \bullet \) (b − a) A1
= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a A1
= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\) A1
\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b A1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} – \) 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
[4 marks]
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\) R1(M1)
Note: Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
a \( \bullet \) b \( = 0\) A1
hence OACB is a rectangle (a and b both non-zero)
with adjacent sides at right angles R1
Note: Award R1(M1)A0R1 if the dot product has not been used.
[4 marks]
Examiners report
In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.
In part (b), some very well structured proofs were offered by a small number of candidates.
Question
OACB is a parallelogram with \(\overrightarrow {{\text{OA}}} = \) a and \(\overrightarrow {{\text{OB}}} = \) b, where a and b are non-zero vectors.
Show that
(i) \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);
(ii) \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.
Answer/Explanation
Markscheme
METHOD 1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}} \)
= (a + b) \( \bullet \) (a + b) A1
= a \( \bullet \) a + a \( \bullet \) b + b \( \bullet \) a + b \( \bullet \) b A1
= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\) A1
\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = – \)(a \( \bullet \) b) A1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
(ii) METHOD 1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}} \)
= (b − a) \( \bullet \) (b − a) A1
= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a A1
= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\) A1
\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b A1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} – \) 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
[4 marks]
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\) R1(M1)
Note: Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
a \( \bullet \) b \( = 0\) A1
hence OACB is a rectangle (a and b both non-zero)
with adjacent sides at right angles R1
Note: Award R1(M1)A0R1 if the dot product has not been used.
[4 marks]
Examiners report
In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.
In part (b), some very well structured proofs were offered by a small number of candidates.