Fundamental Relation of Direction Cosines

If a line makes angles α, β, γ with the positive x, y, z axes respectively, the direction cosines are l = cos α, m = cos β, n = cos γ, and they always satisfy l²+m²+n²=1.

Let a line make angles $\alpha$, $\beta$, $\gamma$ with the positive x, y, z axes. The direction cosines are:

$$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma$$

Consider a unit vector $\hat{u}$ along the line. Its components along the axes are exactly its projections:

$$\hat{u} = (l, m, n) = (\cos\alpha, \cos\beta, \cos\gamma)$$

Since $\hat{u}$ is a unit vector:

$$|\hat{u}|^2 = l^2+m^2+n^2 = 1$$

This is the key constraint: Direction cosines are never independent — fixing two determines the third (up to sign).

Angles $\alpha$, $\beta$, $\gamma$ are not arbitrary: For example, $\alpha = \beta = \gamma$ gives $\cos\alpha = 1/\sqrt{3}$, so $\alpha = \beta = \gamma \approx 54.7°$. One cannot have $\alpha = \beta = \gamma = 45°$ since $3\cos^2(45°) = 3/2 \neq 1$.

Two direction cosine sets per line: A line has two directions (forward and backward). If $(l, m, n)$ are the DCs of one direction, $(-l, -m, -n)$ are the DCs of the other. Both satisfy $l^2+m^2+n^2=1$.