Given any two non-zero vectors a and b , show that \(|\)a \( \times \) b\({|^2}\) = \(|\)a\({|^2}\)\(|\)b\({|^2}\) – (a \( \cdot \) b)\(^2\).

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]

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.