Kuzee
Vector Algebra and Analytic Geometry of Space

Positional Relationships

The Positional Relationship between Two Lines

Let the direction vectors of the two lines be s⃗1=(l1,m1,n1)\vec{s}_1 = (l_1, m_1, n_1) and s⃗2=(l2,m2,n2)\vec{s}_2 = (l_2, m_2, n_2):

Parallel

The condition for two lines to be parallel

s⃗1∥s⃗2  ⟺  l1l2=m1m2=n1n2\vec{s}_1 \parallel \vec{s}_2 \iff \frac{l_1}{l_2} = \frac{m_1}{m_2} = \frac{n_1}{n_2}

Perpendicular

The condition for two lines to be perpendicular

s⃗1⋅s⃗2=0  ⟺  l1l2+m1m2+n1n2=0\vec{s}_1 \cdot \vec{s}_2 = 0 \iff l_1l_2 + m_1m_2 + n_1n_2 = 0

Intersecting

Two lines that are coplanar and not parallel intersect.

Skew

The condition for two lines to be skew

The two lines are not coplanar, that is, there is no common perpendicular perpendicular to both lines.

The Positional Relationship between a Line and a Plane

Let the direction vector of the line be s⃗=(l,m,n)\vec{s} = (l, m, n) and the normal vector of the plane be n⃗=(A,B,C)\vec{n} = (A, B, C):

Parallel

The condition for a line to be parallel to a plane

s⃗⋅n⃗=0  ⟺  Al+Bm+Cn=0\vec{s} \cdot \vec{n} = 0 \iff Al + Bm + Cn = 0

Perpendicular

The condition for a line to be perpendicular to a plane

s⃗∥n⃗  ⟺  lA=mB=nC\vec{s} \parallel \vec{n} \iff \frac{l}{A} = \frac{m}{B} = \frac{n}{C}

Intersecting

The condition for a line to intersect a plane

s⃗⋅n⃗≠0\vec{s} \cdot \vec{n} \neq 0

The Positional Relationship between Two Planes

Let the normal vectors of the two planes be n⃗1=(A1,B1,C1)\vec{n}_1 = (A_1, B_1, C_1) and n⃗2=(A2,B2,C2)\vec{n}_2 = (A_2, B_2, C_2):

Parallel

The condition for two planes to be parallel

n⃗1∥n⃗2  ⟺  A1A2=B1B2=C1C2\vec{n}_1 \parallel \vec{n}_2 \iff \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}

Perpendicular

The condition for two planes to be perpendicular

n⃗1⋅n⃗2=0  ⟺  A1A2+B1B2+C1C2=0\vec{n}_1 \cdot \vec{n}_2 = 0 \iff A_1A_2 + B_1B_2 + C_1C_2 = 0

Intersecting

The condition for two planes to intersect

n⃗1×n⃗2≠0⃗\vec{n}_1 \times \vec{n}_2 \neq \vec{0}

A Summary of the Judging Methods

The Vector Method

  • parallel: the direction vectors are parallel
  • perpendicular: the direction vectors are perpendicular (the scalar product is zero)
  • intersecting: neither parallel nor perpendicular

The Coordinate Method

Judge the positional relationship by comparing the ratios of the coefficients of the direction vectors or the normal vectors.

Exercises

Exercise 1

Judge the positional relationship between the line x−11=y−22=z−33\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and the plane x+y+z=0x + y + z = 0.

the positional relationship between a line and a planethe scalar producttesting for parallelism

Approach:

Compute the scalar product of the direction vector of the line and the normal vector of the plane.

Steps:

  1. The direction vector of the line: (1,2,3)(1, 2, 3)
  2. The normal vector of the plane: (1,1,1)(1, 1, 1)
  3. (1,2,3)⋅(1,1,1)=1+2+3=6≠0(1, 2, 3) \cdot (1, 1, 1) = 1 + 2 + 3 = 6 \neq 0

Answer: the line intersects the plane.

Exercise 2

Judge the positional relationship between the two planes x+y+z=1x + y + z = 1 and x−y+z=2x - y + z = 2.

the positional relationship between two planesthe normal vectorscomputing the angle

Approach:

Compare the ratios of the normal vectors of the two planes, or compute the angle between them.

Steps:

  1. The normal vector of the first plane: (1,1,1)(1, 1, 1)
  2. The normal vector of the second plane: (1,−1,1)(1, -1, 1)
  3. The ratios are not equal, and the scalar product is not zero, so they intersect

Answer: the two planes intersect.

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
s⃗1,s⃗2\vec{s}_1, \vec{s}_2Direction vectordirection vectorsThe direction vectors of the lines
n⃗,n⃗1,n⃗2\vec{n}, \vec{n}_1, \vec{n}_2Normal vectornormal vectorsThe normal vectors of the planes
∥\parallelParallel symbolparallelThe relation of parallelism
⋅\cdotScalar productdot productThe scalar product of vectors
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