MathematicsGrade 11–13

Relative positions and distances in space

Do the lines meet, run parallel or pass each other by? In a 3D space you can turn, intersections, angles and distances become visible.

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What is it about?

Lines are written in parametric form g ⁣:x⃗=a⃗+t⋅u⃗g\colon \vec x = \vec a + t \cdot \vec u, planes in coordinate form E ⁣:n1x1+n2x2+n3x3=cE\colon n_1x_1 + n_2x_2 + n_3x_3 = c with the normal vector n⃗\vec n.

Two lines: if the directions are parallel, the lines are identical or parallel; otherwise they either intersect (the system a⃗+tu⃗=b⃗+sv⃗\vec a + t\vec u = \vec b + s\vec v has a solution) or are skew. The angle between lines satisfies cos⁡φ=∣u⃗⋅v⃗∣∣u⃗∣∣v⃗∣\cos\varphi = \frac{|\vec u \cdot \vec v|}{|\vec u||\vec v|}; skew lines have the distance d=∣(b⃗−a⃗)⋅(u⃗×v⃗)∣∣u⃗×v⃗∣d = \frac{|(\vec b - \vec a)\cdot(\vec u \times \vec v)|}{|\vec u \times \vec v|}.

Line and plane: if n⃗⋅u⃗=0\vec n \cdot \vec u = 0 the line lies in EE or is parallel to it; otherwise there is one point of intersection, and sin⁡φ=∣n⃗⋅u⃗∣∣n⃗∣∣u⃗∣\sin\varphi = \frac{|\vec n \cdot \vec u|}{|\vec n||\vec u|}.

Two planes are parallel (or identical) when their normals are parallel; otherwise they meet in a line with direction n⃗×m⃗\vec n \times \vec m.

Distance point – plane (Hesse normal form): d(P;E)=∣n1p1+n2p2+n3p3−c∣∣n⃗∣d(P; E) = \frac{|n_1p_1 + n_2p_2 + n_3p_3 - c|}{|\vec n|}. The foot FF lies on the perpendicular x⃗=p⃗+tn⃗\vec x = \vec p + t\vec n; going the same distance again gives the mirror point P′P'.

Try it

  1. Play the animation for the skew lines and watch the path light up in the decision tree. Then use “View along g”. Load example
  2. Drag the tip of v⃗\vec v until hh meets gg.
  3. Nearly parallel lines meet far away at a very small angle. Small angle
  4. View two intersecting planes along their line of intersection: the angle appears in true size. Line of intersection
  5. Drop the perpendicular, switch on the mirror point and drag PP into the plane. Distance

Tasks

Task 1: Intersecting lines

Find the point and angle of intersection of g ⁣:x⃗=(1,0,0)+t(1,1,0)g\colon \vec x = (1, 0, 0) + t(1, 1, 0) and h ⁣:x⃗=(0,1,0)+s(1,−1,0)h\colon \vec x = (0, 1, 0) + s(1, -1, 0). Load task

Show solution

1+t=s1 + t = s, t=1−st = 1 - s give t=0t = 0, s=1s = 1, so S(1,0,0)S(1, 0, 0). Since u⃗⋅v⃗=0\vec u \cdot \vec v = 0, the angle is 90°90°. Show in the simulation

Task 2: Point of intersection with a plane

Intersect g ⁣:x⃗=(0,0,5)+t(1,1,−1)g\colon \vec x = (0, 0, 5) + t(1, 1, -1) with E ⁣:x1+x2+x3=2E\colon x_1 + x_2 + x_3 = 2. Load task

Show solution

5+t=25 + t = 2, so t=−3t = -3 and D(−3,−3,8)D(-3, -3, 8); sin⁡φ=13\sin\varphi = \frac13, φ≈19.5°\varphi \approx 19.5°. Show in the simulation

Task 3: Distance point – plane

Find d(P;E)d(P; E), the foot FF and the mirror point for P(1,2,3)P(1, 2, 3) and E ⁣:2x1−x2+2x3=3E\colon 2x_1 - x_2 + 2x_3 = 3. Load task

Show solution

d=∣2−2+6−3∣3=1d = \frac{|2 - 2 + 6 - 3|}{3} = 1, t=−13t = -\frac13, F(13,73,73)F\left(\frac13, \frac73, \frac73\right), P′(−13,83,53)P'\left(-\frac13, \frac83, \frac53\right). Show in the simulation

Task 4: Parallel planes

Find the distance between x1+x2+x3=1x_1 + x_2 + x_3 = 1 and x1+x2+x3=4x_1 + x_2 + x_3 = 4. Load task

Show solution

d=33=3≈1.73d = \frac{3}{\sqrt 3} = \sqrt 3 \approx 1.73. Show in the simulation

Ideas for the classroom

  • Let students predict the path through the decision tree before playing the animation; each question is one step of the calculation.
  • The edge-on views show why distances and angles appear in true size when looking parallel to a plane.
  • Common stumbling blocks:
    • Skew lines look as if they cross in the oblique picture – rotating the view helps.
    • Using cosine instead of sine for the angle between a line and a plane.
    • Using the same parameter name for both lines.
    • Forgetting the absolute value or the division by ∣n⃗∣|\vec n| in the Hesse normal form.