Mathematics

Metric problems in ℝ³

Calculation of angles and distances between points, lines and planes, as well as orthogonal projections.

  • Between two lines: Let rr and ss be two lines, u⃗\vec{u} and v⃗\vec{v} their vectors:
rs^=arccos⁡∣u⃗⋅v⃗∣∣u⃗∣⋅∣v⃗∣\widehat{rs} = \arccos \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}| \cdot |\vec{v}|}
  • Between two planes: π\pi and π′\pi', and let n⃗π\vec{n}_\pi y n⃗π′\vec{n}_{\pi'} be their normal vectors.
ππ′^=arccos⁡∣n⃗π⋅n⃗π′∣∣n⃗π∣⋅∣n⃗π′∣\widehat{\pi\pi'} = \arccos \frac{|\vec{n}_\pi \cdot \vec{n}_{\pi'}|}{|\vec{n}_\pi| \cdot |\vec{n}_{\pi'}|}
  • Between a line r with vector u⃗\vec{u} and a plane with normal vector n⃗π\vec{n}_\pi:
rπ^=arcsin⁡∣u⃗⋅n⃗π∣∣u⃗∣⋅∣n⃗π∣\widehat{r\pi} = \arcsin \frac{|\vec{u} \cdot \vec{n}_\pi|}{|\vec{u}| \cdot |\vec{n}_\pi|}
  • From a point to a line: Let PP be a point and rr a line with vector u⃗\vec{u} and AA any point on the line.
d(P,r)=∣AP⃗×u⃗∣∣u⃗∣d(P, r) = \frac{|\vec{AP} \times \vec{u}|}{|\vec{u}|}
  • From a point P(x0,y0,z0)P(x_0, y_0, z_0) to a plane π\pi in implicit form:
d(P,π)=∣Ax0+By0+Cz0+D∣A2+B2+C2d(P, \pi) = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
  • Between two parallel lines: It will be the distance from any point on one of them to the other line.
  • Between two parallel planes: given in implicit form
d(π,π′)=∣D−D′∣A2+B2+C2d(\pi, \pi') = \frac{|D - D'|}{\sqrt{A^2 + B^2 + C^2}}
  • From a line to a parallel plane: It will be the distance from any point on the line to the plane.
  • Between two skew lines r and s: Let u⃗\vec{u} and v⃗\vec{v} be the vectors and AA and BB the points of rr and ss respectively.
d(r,s)=∣[AB⃗,u⃗,v⃗]∣∣u⃗×v⃗∣d(r, s) = \frac{|[\vec{AB}, \vec{u}, \vec{v}]|}{|\vec{u} \times \vec{v}|}
Find the line r, perpendicular to the plane π\pi, passing through the exterior point P=(x0,y0,z0)P = (x_0, y_0, z_0):
Let the plane be π≡Ax+By+Cz+D=0\pi \equiv Ax + By + Cz + D = 0
It is satisfied that the normal vector of the plane, n⃗\vec{n}, and the vector of the line u⃗\vec{u} are the same:
u⃗=n⃗=(A,B,C)\vec{u} = \vec{n} = (A,B,C)

The parametric equations of the line will be:
{x=x0+λAy=y0+λBz=z0+λC\begin{cases} x = x_0 + \lambda A \\ y = y_0 + \lambda B \\ z = z_0 + \lambda C \end{cases}
Line perpendicular to a plane
Find the plane π\pi, perpendicular to the line r, passing through P=(x0,y0,z0)P = (x_0, y_0, z_0):
This is the opposite problem to the previous one. We obtain the direction vector of the line u⃗=(A,B,C)\vec{u} = (A,B,C).
Since the normal vector of the plane, n⃗\vec{n}, and the vector of the line u⃗\vec{u} are equal, the plane we are looking for is:
π≡Ax+By+Cz+D=0\pi \equiv Ax + By + Cz + D = 0

Where the independent term D remains to be determined. We find it by applying the point D=−(Ax0+By0+Cz0)D = -(Ax_0 + By_0 + Cz_0)
Plane perpendicular to a line
Orthogonal projection of a point P onto a line r: the required point Q is the shadow that P casts on the line. Method:
  • The point Q belongs to the line. Using the parametric equations, Q will have the form:
    Q=(x0+λux,y0+λuy,z0+λuz)Q = (x_0 + \lambda u_x, y_0 + \lambda u_y, z_0 + \lambda u_z)
  • We find the vector PQ⃗\vec{PQ}
  • The vector PQ⃗\vec{PQ} and the direction vector u⃗\vec{u} of the line are perpendicular. We find λ\lambda by applying: PQ⃗⋅u⃗=0\vec{PQ} \cdot \vec{u} = 0
Orthogonal projection of point onto line
Orthogonal projection of a point P onto a plane π\pi: We will follow the next method to find it:
  • We find the equation of the line perpendicular to π\pi passing through P
  • The point Q is the intersection of the found line with π\pi
Orthogonal projection of point onto plane
Symmetric point of a point P with respect to another point Q:
The symmetric point P' we are looking for satisfies:
PQ⃗=QP′⃗\vec{PQ} = \vec{QP'}
Symmetric point with respect to another point
Example: Symmetric of P(3,2,-1) with respect to Q(2,4,0)
We look for P′=(x,y,z)P' = (x,y,z) such that PQ⃗=QP′⃗\vec{PQ} = \vec{QP'}
Then: (2−3,4−2,0+1)=(x−2,y−4,z)(2-3, 4-2, 0+1) = (x-2, y-4, z)
We obtain: P′=(1,6,1)P' = (1,6,1)
Symmetric point of a point P with respect to a line r:
  • We find the projection Q, of point P onto the line r
  • We find the symmetric of P with respect to Q
Symmetric point with respect to a line
Symmetric point of a point P with respect to a plane π\pi:
  • We find the projection Q, of point P onto the plane π\pi
  • We find the symmetric of P with respect to Q
Symmetric point with respect to a plane
Line perpendicular to two skew lines r and s:
Let u⃗\vec{u} and v⃗\vec{v} be the vectors and A and B the points of r and s
  • We find a common perpendicular vector: w⃗=u⃗×v⃗\vec{w} = \vec{u} \times \vec{v}
  • We find the plane π1\pi_1, containing r, with u⃗,w⃗,A\vec{u}, \vec{w}, A
  • We find the plane π2\pi_2, containing s, with v⃗,w⃗,B\vec{v}, \vec{w}, B
  • The required line is the intersection of π1\pi_1 and π2\pi_2
Line that intersects two skew lines r and s, and passes through a point P:
Let u⃗\vec{u} and v⃗\vec{v} be the vectors and A and B the points of r and s. The required line is the intersection of the following planes:
  • Plane π1\pi_1 containing r and P. It is found with u⃗,AP⃗,A\vec{u}, \vec{AP}, A
  • Plane π2\pi_2 containing s and P. It is found with v⃗,BP⃗,B\vec{v}, \vec{BP}, B