Sine, not cosine: The dot product gave you the cosine: v·w = |v||w|cos θ. The cross product gives you the sine.
The length: |v×w| = |v||w| sin θ, where θ is the angle between v and w (0 ≤ θ ≤ 180°).
Parallel (θ = 0)
- sin θ = 0, so |v×w| = 0.
- This confirms that parallel vectors have a zero cross product.
Perpendicular (θ = 90°)
- sin θ = 1.
- The length is biggest.
IB-style question — length two ways
Vectors v and w have |v| = 5 and |w| = 4, with an angle of 30° between them.
Find |v×w|.
Step by step
- Use the length formula with the given magnitudes and angle.
- Evaluate (sin 30° = ½).
Final answer
|v×w| = 10.
IB-style question — find the angle from the length
For vectors a and b, |a| = 6, |b| = 2 and |a×b| = 6.
Find the acute angle θ between them.
Step by step
- Rearrange the length formula for sin θ.
- Take the inverse sine (acute angle).
Final answer
θ = 30°.
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|v×w| IS the parallelogram area: Picture a parallelogram with sides v and w. Its area is exactly |v×w|.
Why the area is |v×w|
Base × height
Area = |v| × (|w| sin θ).
The height
The height is the slanted side |w| projected at right angles to the base.
Match it
That's exactly |v||w| sin θ = |v×w|.
Areas
- Parallelogram area = |v×w|.
- A triangle is half of that parallelogram, so Triangle area = ½|v×w|.
The recipe for a triangle ABC: 1. Form two edge vectors from one corner, e.g. AB = B − A and AC = C − A.
2. Compute AB×AC.
3. Take its length, then halve it: area = ½|AB×AC|.
This beats Heron's formula or ½ab sin C because you never have to find an angle.
IB-style question — area of a triangle in 3D
Triangle ABC has A(1, 0, 1), B(3, 2, 1) and C(2, 0, 4).
Find the area of triangle ABC.
Step by step
- Edge vectors from A.
- Cross product. i: (2)(3)−(0)(0)=6. j: (0)(1)−(2)(3)=−6. k: (2)(0)−(2)(1)=−2.
- Length of the cross product.
- Triangle area is half this.
Final answer
Area = ½√76 = √19 ≈ 4.36 (square units).