Fundamentals of Three-Dimensional Computational Geometry

Many concepts in three-dimensional geometry are shared with two-dimensional geometry, and the same methods used for two-dimensional problems can be applied to three-dimensional problems.

Basic concepts

The concepts of points, vectors, and lines are similar to those in two-dimensional geometry and will not be expanded on here.

Planes

A plane can be represented by a point P0(x0,y0,z0)P_0(x_0,y_0,z_0) on the plane and a normal vector (that is, a vector perpendicular to the plane) n\boldsymbol{n}.

Because n\boldsymbol{n} is perpendicular to the plane, n\boldsymbol{n} is perpendicular to every line in the plane. In other words, if n=(A,B,C)\boldsymbol{n}=(A,B,C), every point P(x,y,z)P(x,y,z) on the plane satisfies n⋅PP0→=0\boldsymbol{n} \cdot \overrightarrow{PP_0} = 0.

By the definition of the dot product, this is equivalent to:

A(x−x0)+B(y−y0)+C(z−z0)=0A(x-x_0)+B(y-y_0)+C(z-z_0)=0

Rearranging gives:

Ax+By+Cz−(Ax0+By0+Cz0)=0Ax+By+Cz-(Ax_0+By_0+Cz_0)=0

Let D=−(Ax0+By0+Cz0)D=-(Ax_0+By_0+Cz_0). The equation then becomes Ax+By+Cz+D=0Ax+By+Cz+D=0, which is called the general form of the plane equation.

Basic operations

Angles between lines and planes

Using spatial vectors, angles between lines and planes in space can be found quickly.

For two skew lines aa and bb, take a point PP in space and draw a′∥aa' \parallel a and b′∥b′b' \parallel b'. The acute or right angle between a′a' and b′b' is called the angle between the skew lines aa and bb.

For a line aa and a plane α\alpha, suppose that aa intersects α\alpha at AA. Through a point PP on aa, draw the perpendicular to plane α\alpha, meeting α\alpha at OO. The complement of the angle between aa and POPO is called the angle between a line and a plane. In particular, if a∥αa \parallel \alpha or a⊂αa \subset \alpha, their angle is 0∘0^\circ.

For two planes α\alpha and β\beta, their angle is defined as the angle between two lines a,ba,b perpendicular to their intersection line ll, where a⊂αa \subset \alpha and b⊂βb \subset \beta.

Definition of the angle between two lines and its necessary and sufficient conditions

  • The angle between the direction vectors of two lines is called the angle between the two lines.

Given lines l1,l2l_1, l_2 with direction vectors s1(m1,n1,p1)s_1 (m_1, n_1, p_1) and s2(m2,n2,p2)s_2 (m_2, n_2, p_2), and letting φ\varphi be the angle between the lines, we have cos⁡φ=∣m1m2+n1n2+p1p2∣m12+n12+p12m22+n22+p22\cos \varphi = \dfrac{\left | m_1m_2+n_1n_2+p_1p_2 \right |}{\sqrt{m_1^2+n_1^2+p_1^2}\sqrt{m_2^2+n_2^2+p_2^2}}.

  • l1⊥l2  ⟺  m1m2+n1n2+p1p2=0l_1 \perp l_2 \iff m_1m_2 + n_1n_2 + p_1p_2 = 0

  • l1∥l2  ⟺  m1m2=n1n2=p1p2l_1 \parallel l_2 \iff \dfrac{m_1}{m_2} = \dfrac{n_1}{n_2} = \dfrac{p_1}{p_2}.

Angles between three-dimensional vectors and planes

When a line is not perpendicular to a plane, the angle φ\varphi between the line and its projection onto the plane (φ∈[0,π2]\varphi \in [0, \frac{\pi}{2}]) is called the angle between the line and the plane.

Let the direction vector of a line be s(m,n,p)s(m, n, p) and the normal vector of a plane be f(a,b,c)f(a, b, c). Then:

  • Sine of the angle: sin⁡φ=∣am+bn+cp∣a2+b2+c2m2+n2+p2\sin\varphi = \dfrac{\left | am + bn + cp \right |}{\sqrt{a^2+b^2+c^2}\sqrt{m^2+n^2+p^2}}

  • The line is parallel to the plane   ⟺  am+bn+cp=0\iff am+bn+cp = 0

  • The line is perpendicular to the plane   ⟺  am=bn=cp\iff \dfrac{a}{m} = \dfrac{b}{n} = \dfrac{c}{p}

Distance from a point to a plane

Intersection of a line and a plane

Simply solve the line and plane equations simultaneously.

Theorems of solid geometry

Three-sine theorem

Let the dihedral angle M-AB-NM-AB-N have measure α\alpha. On plane MM, let ray ACAC form an angle β\beta with edge ABAB and an angle γ\gamma with plane NN. Then sin⁡γ=sin⁡α⋅sin⁡β\sin\gamma = \sin\alpha\cdot\sin\beta.

Three-cosine theorem

Let OO be a point on a plane, and let the projection onto the plane of a line BOBO through a point BB outside the plane be AOAO. Let OCOC be a line in the plane. Then the cosines of ∠COB,∠AOC,∠AOB\angle COB,\angle AOC,\angle AOB satisfy cos⁡∠BOC=cos⁡∠AOB⋅cos⁡∠AOC\cos\angle BOC=\cos\angle AOB\cdot\cos\angle AOC (both ∠AOC\angle AOC and ∠AOB\angle AOB must be acute).

References