IBDP Maths analysis and approaches Topic: AHL 3.16 Properties of the vector product HL Paper 1

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]

a.

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]

b.
Answer/Explanation

Markscheme

a \( \times \) b \( =  – 12\)i \( – {\text{ }}2\)j \( – {\text{ }}3\)k     (M1)A1

[2 marks]

a.

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]

b.
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