Coplanar Vectors

An Intuitive Understanding of Coplanarity

Coplanar vectors

Three vectors are coplanar if, after being freely translated, they lie in the same plane.

The Necessary and Sufficient Condition for Coplanarity

The necessary and sufficient condition for coplanarity

Three vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are coplanar if and only if there exist real numbers k,l,mk, l, m, not all zero, such that:

ka⃗+lb⃗+mc⃗=0⃗k\vec{a} + l\vec{b} + m\vec{c} = \vec{0}
Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
k,l,mk, l, mReal numbersReal numbersThe coefficients of the linear combination

Methods for Testing Coplanarity

  1. The scalar triple product method: compute a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c}); if it is zero, they are coplanar
  2. The linear-combination method: check whether there exist k,l,mk, l, m not all zero with ka⃗+lb⃗+mc⃗=0⃗k\vec{a} + l\vec{b} + m\vec{c} = \vec{0}
  3. The determinant method: compute the determinant formed by the coordinates of the three vectors; if it is zero, they are coplanar
  4. The volume method: if the parallelepiped spanned by the three vectors has zero volume, they are coplanar

Worked Examples and Applications

A positive example: coplanar vectors

Example 1: decide whether the vectors a⃗=(1,0,0)\vec{a} = (1, 0, 0), b⃗=(0,1,0)\vec{b} = (0, 1, 0) and c⃗=(1,1,0)\vec{c} = (1, 1, 0) are coplanar.

Solution:

  1. The scalar triple product: a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c})

    • b⃗×c⃗=(0,1,0)×(1,1,0)=(0,0,−1)=(0,0,−1)\vec{b} \times \vec{c} = (0, 1, 0) \times (1, 1, 0) = (0, 0, -1) = (0, 0, -1)
    • a⃗⋅(0,0,−1)=1⋅0+0⋅0+0⋅(−1)=0\vec{a} \cdot (0, 0, -1) = 1 \cdot 0 + 0 \cdot 0 + 0 \cdot (-1) = 0
  2. Or the linear combination: look for k,l,mk, l, m with k(1,0,0)+l(0,1,0)+m(1,1,0)=(0,0,0)k(1,0,0) + l(0,1,0) + m(1,1,0) = (0,0,0)

    • the system: k+m=0k + m = 0, l+m=0l + m = 0, 0=00 = 0
    • solving gives k=1k = 1, l=1l = 1, m=−1m = -1 (the coefficients are not all zero)

A negative example: non-coplanar vectors

Example 2: decide whether the vectors a⃗=(1,0,0)\vec{a} = (1, 0, 0), b⃗=(0,1,0)\vec{b} = (0, 1, 0) and c⃗=(0,0,1)\vec{c} = (0, 0, 1) are coplanar.

Solution:

  1. The scalar triple product: a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c})

    • b⃗×c⃗=(0,1,0)×(0,0,1)=(1,0,0)\vec{b} \times \vec{c} = (0, 1, 0) \times (0, 0, 1) = (1, 0, 0)
    • a⃗⋅(1,0,0)=1⋅1+0⋅0+0⋅0=1≠0\vec{a} \cdot (1, 0, 0) = 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 ≠ 0
  2. Or the determinant:

    ∣100010001∣=1≠0\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 ≠ 0

An example of linear dependence

Example 3: decide whether the vectors a⃗=(1,2,3)\vec{a} = (1, 2, 3), b⃗=(2,4,6)\vec{b} = (2, 4, 6) and c⃗=(3,6,9)\vec{c} = (3, 6, 9) are coplanar.

Solution:

  1. Observe the relation: b⃗=2a⃗\vec{b} = 2\vec{a}, c⃗=3a⃗\vec{c} = 3\vec{a}
  2. The linear combination: 1a⃗+(−2)b⃗+1c⃗=(0,0,0)1\vec{a} + (-2)\vec{b} + 1\vec{c} = (0,0,0)
  3. The coefficients (1,−2,1)(1, -2, 1) are not all zero

Application Scenarios

Applications in spatial geometry

  1. Equations of planes: the plane determined by three points in space can be described by using coplanar vectors to represent the position vector of any point on the plane
  2. Linear dependence: coplanarity of vectors is an important concept in studying linear dependence
  3. Solid geometry: coplanar vectors play an important role in solving problems in spatial geometry

A reminder of common misconceptions

Practice

Exercise 1

Decide whether the vectors a⃗=(1,1,0)\vec{a} = (1, 1, 0), b⃗=(1,0,1)\vec{b} = (1, 0, 1) and c⃗=(0,1,1)\vec{c} = (0, 1, 1) are coplanar.

coplanar vectorsthe scalar triple productthe determinant

Approach:

Use the scalar triple product or the determinant to decide whether the three vectors are coplanar.

Steps:

  1. Compute the scalar triple product:

    • b⃗×c⃗=(1,0,1)×(0,1,1)=(0−1,−(1−0),1−0)=(−1,−1,1)\vec{b} \times \vec{c} = (1, 0, 1) \times (0, 1, 1) = (0-1, -(1-0), 1-0) = (-1, -1, 1)
    • a⃗⋅(−1,−1,1)=1⋅(−1)+1⋅(−1)+0⋅1=−2≠0\vec{a} \cdot (-1, -1, 1) = 1\cdot(-1) + 1\cdot(-1) + 0\cdot1 = -2 ≠ 0
  2. Or compute the determinant:

    ∣110101011∣=1(0−1)−1(1−0)+0(1−0)=−1−1+0=−2≠0\begin{vmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = 1(0-1) - 1(1-0) + 0(1-0) = -1 - 1 + 0 = -2 ≠ 0

Answer: The vectors a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} are not coplanar, because the scalar triple product is not zero.

Exercise 2

Decide whether the vectors a⃗=(1,3,2)\vec{a} = (1, 3, 2), b⃗=(2,6,4)\vec{b} = (2, 6, 4) and c⃗=(3,9,6)\vec{c} = (3, 9, 6) are coplanar.

coplanar vectorslinear dependencethree-dimensional space

Approach:

Check whether there exist real numbers, not all zero, making the linear combination the zero vector.

Steps:

  1. Observe the relation between the coordinates: b⃗=2a⃗\vec{b} = 2\vec{a}, c⃗=3a⃗\vec{c} = 3\vec{a}
  2. The linear combination: 1a⃗+(−2)b⃗+1c⃗=(0,0,0)1\vec{a} + (-2)\vec{b} + 1\vec{c} = (0,0,0)
  3. The coefficients (1,−2,1)(1, -2, 1) are not all zero

Answer: The three vectors are coplanar, because they are linearly dependent.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}VectorVector a, b, cGeneral vectors
k,l,mk, l, mReal numbersReal numbersThe coefficients of the linear combination

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