Key Idea: The vector (cross) product v×w gives a vector at right angles to both v and w. Its length gives areas of parallelograms and triangles. Do it by hand on Paper 1; the GDC can do it on Paper 2.
The cross product by components
- Output is a vector (the dot product gives a number).
- Work it out with the 3×3 determinant. Mind the minus on j.
- w×v = −(v×w).
- Check: v·(v×w) = 0.
| Quantity | Formula |
|---|---|
| Length | |v×w| = |v||w| sin θ |
| Parallelogram area | |v×w| |
| Triangle area | ½|v×w| |
| Without θ | |v×w|² = |v|²|w|² − (v·w)² |
For a triangle ABC, cross AB and AC from the same corner, then halve.
(1, 2, −1)×(2, 0, 3)? (6, −5, −4).
How do you check that answer? Dot it with (1, 2, −1): 6 − 10 + 4 = 0.
|v| = 3, |w| = 5, v·w = 9. Triangle area? |v×w|² = 225 − 81 = 144, so the area is ½ × 12 = 6.
Your answer has the opposite signs to the markscheme. Why? You crossed in the other order. Both are valid normals.