A **Spanning Set** is the smallest set of vectors where any vector in a certain space
can be represented as a **Linear Combination** of the spanning set vectors.

The **Spanning Set** can span a line (1-dimension), a plane (2-dimension) or a space
(3-dimensions).

Look how the **Standard Basis** unit vectors can be used to create **Spanning Set**:

Spanning Set | Line | Spanning Set | Plane | Spanning Set | Space |

\(\hat{\imath}\) | x-dir | \(\hat{\imath}\), \(\hat{\jmath}\) | xy-plane | \(\hat{\imath}\), \(\hat{\jmath}\), \(\hat{k}\) | 3D space |

\(\hat{\jmath}\) | y-dir | \(\hat{\imath}\), \(\hat{k}\) | xz-plane | ||

\(\hat{k}\) | z-dir | \(\hat{\jmath}\), \(\hat{k}\) | yz-plane |

Given a spanning set of just \(\hat{\imath}\), the only vectors we can make using linear combnations are vectors along the x-axis. Therefore, it spans a line.

\(2\hat{\imath}\),\(-1.5\hat{\imath}\), etc.

Given a spanning set of \(\hat{\jmath}\), \(\hat{k}\), the only vectors we can make using linear combinations are vectors in the yz-plane. Therefore, they span a plane.

\(2\hat{\jmath} + \hat{k}\), \(-2\hat{\jmath} + \hat{k}\), etc.

Given a spanning set of \(\hat{\imath}\), \(\hat{\jmath}\), \(\hat{k}\), we can make any vector in 3D space using linear combinations. Therefore, they space a space.

\(\hat{\imath} + \hat{\jmath} - 0.5\hat{k} \), etc.

What is the minimum number of vectors to span a plane?

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What is the minimum number of vectors to span a plane from \( \left<10, 4 \right> \),
\( \left<1, 0 \right> \) and \( \left<0, 1 \right> \)?

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Let's see a few more examples:

The lines are Coplanar. They both lie on a plane. |
The lines are Coplanar. They all lie on a plane. |
The lines are non-Coplanar. They all cannot lie on the same plane. |

Spans a plane. |
Span a plane. |
Span a space. |

What do the following vectors span? \( \vec{u} = \left<3,-1,4 \right>,
\vec{v} = \left<6,-4,-8\right>, \vec{w} = \left<7,-3,4 \right>\)

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