Calculating using a determinant and testing for collinearity

A practical notation using determinants

To avoid sign errors, the vector product is often expressed using 2×2 mini-determinants. If u = (x1, y1, z1) and v = (x2, y2, z2):

u ^ v = ( |y1 z1; y2 z2| , -|x1 z1; x2 z2| , |x1 y1; x2 y2| )

where |a b; c d| = ad - bc denotes the 2×2 determinant. Be careful with the minus sign in front of the second component: this is the most common mistake!

Detailed example

u = (1, 2, 3) and v = (4, 5, 6).

• First component: y1z2 - z1y2 = 26 - 35 = 12 - 15 = -3
• Second component: -(x1z2 - z1x2) = -(16 - 34) = -(6-12) = 6
• Third component: x₁y₂ - y₁x₂ = 15 - 24 = 5 - 8 = -3

Therefore, u ^ v = (-3, 6, -3).

Testing for collinearity

Two non-zero vectors u and v are collinear if and only if u ^ v = (0, 0, 0). This is often quicker than finding a proportionality coefficient, especially when one coordinate is zero.

Example

u = (2, -1, 3) and v = (-4, 2, -6). We can see that v = -2*u, so they are collinear: calculating u ^ v will indeed give (0, 0, 0). You can check this component by component to practise.

Common pitfall

A common mistake is to forget the negative sign in the second component, or to swap u and v during the calculation without realising it, which reverses the sign of the final result. Get into the habit of checking your result using a scalar product: u . (u ^ v) must always equal 0.