Plane Inclination Angles

Summary: §§97–100 of the Appendix on Surfaces. First-order surfaces are exactly planes (§§96–98) — the 3D analogue of straight-line-equation. Given the plane $\alpha x+\beta y+\gamma z=a$, Euler then computes its angles of inclination to all three coordinate planes (§§99–100, figure 142): the tangent of the inclination to the plane $APQ$ is $-\sqrt{{\alpha }^2+{\beta }^2}/\gamma$, with cyclic versions $-\sqrt{{\alpha }^2+{\gamma }^2}/\beta$ and $-\sqrt{{\beta }^2+{\gamma }^2}/\alpha$. These are the direction-cosine computations in elementary 3D analytic geometry.

Sources: appendix4, §§97–100. Figure 142 in figures142-144.

Last updated: 2026-05-12.

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§§96–98 — Planes = first-order surfaces

A surface whose intersection with a plane is always a straight line must be a plane. (source: appendix4, §97)

The argument:

1. Curve analogue: a plane curve whose intersection with every line is a single point is itself a line (Book II, line-curve-intersection-bound converse).
2. By degree-invariance-surfaces §95, every plane section of a first-order surface is a first-order curve, i.e., a straight line.
3. If any convexity or concavity existed, some section would be a curve — contradiction. So the surface is flat.

§98’s algebraic confirmation: starting from $\alpha x+\beta y+\gamma z=a$, the change-of-coordinates-3d §92 substitution has enough freedom to reduce this to $r=f$ (a plane parallel to the $pq$-plane) or even $r=0$ (the $pq$-plane itself).

§99 — Trace in the principal plane (figure 142)

Set $z=0$: the trace is the line $\alpha x+\beta y=a$ in the plane $APQ$. Geometrically this line cuts:

• The axis $AP$ at $AC=a/\alpha$.
• The perpendicular through $A$ at $AB=a/\beta$.

Hence

$$\tan (\angle ACB)=\frac{AB}{AC}\cdot \frac{1}{1}=\frac{\alpha }{\beta },\sin =\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}},\cos =\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}.$$

Extending the ordinate $QP$ to meet the trace at $R$:

$$CR=\frac{x\sqrt{{\alpha }^2+{\beta }^2}}{\beta }-\frac{a\sqrt{{\alpha }^2+{\beta }^2}}{\alpha \beta },PR=\frac{\alpha x-a}{\beta }.$$

§100 — Inclination to the coordinate plane $APQ$

From $Q$ drop $QS\perp BC$ in the plane $APQ$, and join $M$ (on the surface, with $z=QM$) to $S$. The angle $\angle MSQ$ measures the plane’s inclination to $APQ$.

Computing $QR$ in two ways:

$$QR=\frac{\alpha x+\beta y-a}{\beta }=\frac{-\gamma z}{\beta }$$

(using $\alpha x+\beta y+\gamma z=a$). Since $\angle RQS=\angle ACB$:

$$QS=QR\cdot \cos (\angle ACB)=\frac{-\gamma z}{\sqrt{{\alpha }^2+{\beta }^2}}.$$

Hence

$$\boxed{\tan (\angle QSM)=\frac{QM}{QS}=\frac{-\sqrt{{\alpha }^2+{\beta }^2}}{\gamma }.}$$

So

$$\cos (\angle QSM)=\frac{\gamma }{\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}}.$$

Three inclinations by cyclic symmetry

The same calculation with the plane $APR$ or $AQR$ playing the role of “principal plane” gives:

$$\tan (\text{inclination to }APQ)=\frac{-\sqrt{{\alpha }^2+{\beta }^2}}{\gamma },$$ $$\tan (\text{inclination to }APR)=\frac{-\sqrt{{\alpha }^2+{\gamma }^2}}{\beta },$$ $$\tan (\text{inclination to }AQR)=\frac{-\sqrt{{\beta }^2+{\gamma }^2}}{\alpha }.$$

In modern terms: $(\alpha ,\beta ,\gamma )$ is the unnormalized plane normal; the three tangents above are the ratios of horizontal-component magnitude to vertical-component magnitude for each coordinate plane choice.

Cross-references

• §§96–98 lifts straight-line-equation one dimension up.
• The angles here are the analytic substrate of the projection method in oblique-plane-section-method — the inclination angle $\varphi$ in §49’s substitution is precisely $\angle QSM$ for the cutting plane.
• The expressions are the same direction-cosine ratios that appear in $\vec{n}\cdot {\hat{e}}_i/|\vec{n}|$ in modern vector analysis.

Figures

Figures 142–144

Related pages

• appendix-4-on-the-change-of-coordinates
• change-of-coordinates-3d
• degree-invariance-surfaces
• oblique-plane-section-method
• straight-line-equation