Question
Given any two non-zero vectors a and b , show that \(|\)a \( \times \) b\({|^2}\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\).
Answer/Explanation
Markscheme
METHOD 1
Use of \(|\)a \( \times \) b\(|\) = \(|\)a\(|\)\(|\)b\(|\)\(\sin \theta \) (M1)
\(|\)a \( \times \) b\({|^2}\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\sin ^2}\theta \) (A1)
Note: Only one of the first two marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\((1 – {\cos ^2}\theta )\) A1
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\cos ^2}\theta \) (A1)
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(\left( | \right.\)a\(|\)\(|\)b\(|\)\({\left. {\cos \theta } \right)^2}\) (A1)
Note: Only one of the above two A1 marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\) A1
Hence LHS = RHS AG N0
[6 marks]
METHOD 2
Use of a \( \cdot \) b = \(|\)a\(|\)\(|\)b\(|\)\(\cos \theta \) (M1)
\(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(\left( | \right.\)a\(|\)\(|\)b\(|\)\({\left. {\cos \theta } \right)^2}\) (A1)
= \(|\)a\({|^2}\)\(|\)b\({|^2}\) – \(|\)a\({|^2}\)\(|\)b\({|^2}\) \({\cos ^2}\theta \) (A1)
Note: Only one of the above two A1 marks can be implied.
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\((1 – {\cos ^2}\theta )\) A1
= \(|\)a\({|^2}\)\(|\)b\({|^2}\)\({\sin ^2}\theta \) A1
= \(|\)a \( \times \) b\({|^2}\) A1
Hence LHS = RHS AG N0
Notes: Candidates who independently correctly simplify both sides and show that LHS = RHS should be awarded full marks.
If the candidate starts off with expression that they are trying to prove and concludes that \({\sin ^2}\theta = (1 – {\cos ^2}\theta )\) award M1A1A1A1A0A0.
If the candidate uses two general 3D vectors and explicitly finds the expressions correctly award full marks. Use of 2D vectors gains a maximum of 2 marks.
If two specific vectors are used no marks are gained.
[6 marks]
Examiners report
Those candidates who chose to use the trigonometric version of Pythagoras’ Theorem were generally successful, although a minority were unconvincing in their reasoning. Some candidates adopted a full component approach, but often seemed to lose track of what they were trying to prove. A few candidates used 2-dimensional vectors or specific rather than general vectors.
Question
Consider the vectors a \( = \) i \( – {\text{ }}3\)j \( – {\text{ }}2\)k, b \( = – {\text{ }}3\)j \( + {\text{ }}2\)k.
Find a \( \times \) b.[2]
Hence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point \((1,{\text{ }}0,{\text{ }} – 1)\).[3]
Answer/Explanation
Markscheme
a \( \times \) b \( = – 12\)i \( – {\text{ }}2\)j \( – {\text{ }}3\)k (M1)A1
[2 marks]
METHOD 1
\( – 12x – 2y – 3z = d\) M1
\( – 12 \times 1 – 2 \times 0 – 3( – 1) = d\) (M1)
\( \Rightarrow d = – 9\) A1
\( – 12x – 2y – 3z = – 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\)
METHOD 2
\(\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 12} \\ { – 2} \\ { – 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ { – 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 12} \\ { – 2} \\ { – 3} \end{array}} \right)\) M1A1
\( – 12x – 2y – 3z = – 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\) A1
[3 marks]