This document outlines steps to prove that the magnitude of the cross product of two vectors a and b is equal to absinθ, where θ is the angle between the vectors. It involves evaluating the determinants of a×b and a2b2 - a·b2, and showing they are equal to absinθ and a2b2sin2θ respectively. Taking the difference of these expressions and using trigonometric identities reveals the relationship is equivalent to the area of the parallelogram formed by the two vectors.

1. IB MHL 1
Proof of the Magnitude of the Vector (Cross) Product
This structured inquiry will help you to be more:
• Precise and accurate
• Resilient and patient
• Appreciative of the beauty and elegance of mathematics
This is what we are trying to prove:
If 𝜃 is the angle between vectors a and b then
𝒂×𝒃 = 𝒂 𝒃 𝑠𝑖𝑛𝜃 for 0 ≤ 𝜃 ≤ 𝜋
In words the magnitude of the cross product = area of a parallelogram formed by
two vectors
Step 1
Let 𝒂 = 𝑥! 𝒊 + 𝑦! 𝒋 + 𝑧! 𝒌 and 𝒃 = 𝑥! 𝒊 + 𝑦! 𝒋 + 𝑧! 𝒌
Evaluate 𝒂×𝒃 using the determinant matrix:
𝒊 𝒋 𝒌
Step 2
Now evaluate 𝒂×𝒃 2 and expand each of the brackets:
Call this equation number (1)

2. IB MHL 2
Step 3
Now work out 𝒂 2 𝒃 2 and multiply the brackets out.
Call this equation 2:
Step 4
Now let us look at the scalar product squared and expand all brackets
𝒂 ∙ 𝒃 2 =
Write all the squared terms at the beginning and call this equation 3:

3. IB MHL 3
Step 5
Work out equation 2 minus equation 3 and cancel out the common terms.
That is:
𝒂 2 𝒃 2 - 𝒂 ∙ 𝒃 2 =
What do you end up with? Which equation is this?

4. IB MHL 4
What is the relationship that connects equations 1,2 and 3?
Replace the scalar product with 𝒂 ∙ 𝒃 = 𝒂 𝒃 𝑐𝑜𝑠𝜃 and take out 𝒂 2 𝒃 2 as a
common factor.
Use 𝑐𝑜𝑠!
𝜃 + 𝑠𝑖𝑛!
𝜃 = 1
What does this mean geometrically?
𝜃
b a