Phone vectors: three vectors, one plane
About this app
The keypad rows 1-2-3, 4-5-6, 7-8-9 as vectors in space: the game of the closed vector chain finds a − 2b + c = 0, so they are linearly dependent and lie in one plane. Seen edge-on, all three arrows lie on one line. For comparison c = (7, 8, 10) sticks out, the chain no longer closes, det ≠ 0.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
The idea
Three vectors a, b, c in ℝ³ are linearly dependent if r·a + s·b + t·c = 0 has a solution in which not all factors are zero. Intuitively, you can stretch, shrink or reverse them and lay them tip to tail into a closed chain. That works exactly when the three vectors lie in a common plane through the origin. To check it algebraically you solve a homogeneous system of linear equations: if it has infinitely many solutions, the vectors are dependent.
What you see
The three rows of a phone keypad become the vectors a = (1, 2, 3), b = (4, 5, 6) and c = (7, 8, 9). They look randomly chosen, yet they lie in one plane: a − 2b + c = 0. The app shows the common plane, the closed chain made of a, −2b and c, and an edge-on view in which all three arrows appear on one line. The values panel shows the determinant, the status and the factors of the chain.
Things to try
Rotate the scene until you look exactly along the edge of the plane. Then change c to (7, 8, 10): the arrow sticks out of the plane, the determinant is no longer zero and the chain no longer closes. With your own numbers you see how rarely three random vectors lie in one plane. In explain mode, the AI tutor guides you through the app.
Frequently asked questions
- When are three vectors in space linearly dependent?
- Exactly when they lie in a common plane through the origin. Then there are numbers r, s, t, not all zero, with r·a + s·b + t·c = 0. Equivalently, the determinant of the matrix with the three vectors as columns is zero.
- Are (1, 2, 3), (4, 5, 6) and (7, 8, 9) linearly independent?
- No. (1, 2, 3) − 2·(4, 5, 6) + (7, 8, 9) = (0, 0, 0). The system r·a + s·b + t·c = 0 therefore has infinitely many solutions and the vectors lie in one plane, although no two of them are parallel.
- How do you test linear independence with Gaussian elimination?
- Write r₁a₁ + … + rₙaₙ = 0 component by component as a homogeneous linear system and solve it by Gaussian elimination. If only r₁ = … = rₙ = 0 remains, the vectors are independent. If an equation drops out because it is a multiple of another, there are infinitely many solutions and the vectors are dependent.
- What does the Phone Vectors app show?
- It turns the rows of a phone keypad into three vectors in space and shows their common plane, the closed chain a − 2b + c = 0 and an edge-on view. All components can be changed, and the determinant and status update live.
Subject: Linear algebra
More from Linear algebra
- Eigenvectors
- Gaussian elimination step by step
- Linear combination & linear independence (3D)
- Matrix × vector
- Matrix multiplication
- Horse and rail: the dot product as work
- Tesseract — the cube of the fourth dimension (3D)
- Vector workshop
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