|u × v| = 15 for u = (1, 2, 2) and v = (4, −2, 3). The triangle on those same two vectors has area 7.5. A factor of two, and the question will have asked for one of them.
The shaded parallelogram is what the vector product measures. The diagonal splits it into two equal triangles, which is where the factor of two comes from.
The vector product of two vectors is a vector, which is the first thing to fix: the scalar product at 3.13 gave a number, and this one gives three.
For u = (1, 2, 2) and v = (4, −2, 3):
| Component | From | Here |
|---|---|---|
| first | u2v3 − u3v2 | 6 − (−4) = 10 |
| second | u3v1 − u1v3 | 8 − 3 = 5 |
| third | u1v2 − u2v1 | −2 − 8 = −10 |
so u × v = (10, 5, −10).
The middle one runs the other way round. First and third both go "lower index times higher, minus higher times lower"; the second does not. Writing it as u1v3 − u3v1 gives −5 instead of 5, and the resulting vector is not perpendicular to anything useful. It is the single commonest arithmetic error on this sub-topic, and the check below catches it.
The free check: the answer must be perpendicular to both inputs. That is what a vector product is for, so:
u · (u × v) = 0 and v · (u × v) = 0
Two scalar products, ten seconds, and a sign error in any component will break at least one of them. Do this every time until the component pattern is automatic.
Swap the two and every component changes sign:
u × v = (10, 5, −10) but v × u = (−10, −5, 10)
v × w = −(w × v), which is called being anticommutative, and it is the first operation on this course where order matters. Addition does not care; the scalar product does not care; this one does.
Geometrically the two point in opposite directions, both perpendicular to the plane of u and v. Which one you get is decided by the right-hand screw rule: point the fingers of your right hand along the first vector, curl them towards the second, and your thumb gives the direction of the product.
A consequence worth noticing: v × v = 0, the zero vector. Any vector is parallel to itself, and the product of parallel vectors vanishes, which is the mirror image of the perpendicular test for the scalar product. So a zero scalar product means perpendicular, and a zero vector product means parallel.
|v × w| = |v| |w| sin θ
Compare that with |v||w| cos θ for the scalar product and the whole relationship falls out. Where one is largest the other is zero. Drag the slider: at 0° the scalar product is 25 and the cross product is 0; at 90° it is the other way round; at 45° both are 17.68. The sum of their squares is a constant 625, because sin² + cos² = 1.
And |v||w| sin θ is exactly the area formula for a parallelogram: one side, times the other, times the sine of the angle between them. So:
For our u and v: |u × v| = √(100 + 25 + 100) = √225 = 15, so the parallelogram is 15 and the triangle is 7.5. Check against the other formula: |u| = 3, |v| = 5.3852 and the angle is 68.2°, whose sine is 0.9285, and 3 × 5.3852 × 0.9285 = 15.
Read the question twice. "Find the area of the triangle ABC" wants half of the magnitude, and "find the area of the parallelogram ABCD" wants all of it. The two answers differ by a factor of two and both look like plausible areas.
Both machines compute a cross product, and the component pattern is fiddly enough by hand that checking is genuinely worth the keystrokes. The better use is the perpendicularity check, which catches a sign error that a recomputation might repeat.
When you may use it. Analysis Paper 1 is non-calculator, and these questions are nearly always Paper 1 because the components are whole numbers by design. Learn the pattern; use the machine to confirm it while you do.
The mark people lose. Giving the parallelogram's area where the triangle's was asked for. It is right to within a factor of two, which earns nothing, and the working looks flawless. The habit: underline the word "triangle" or "parallelogram" in the question before you start, and write "× ½" next to it if it says triangle.
Throughout: u = (1, 2, 2) and v = (4, −2, 3).
1. Find the first component of u × v.
2. Find the second component of u × v.
3. Find the area of the parallelogram spanned by u and v.
4. Find the area of the triangle with two sides u and v.
5. What does u × v = 0 tell you, for non-zero u and v?
1 markThe vector product, computed correctly.
1 markIts magnitude.
1 markThe halving, if a triangle was asked for.
Three marks, and the third is free if you read the question. It is also the one most often lost, because a correct parallelogram area looks like a finished answer.
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