Linear Algebra - Find intersection of geometric objects

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Find intersection of geometric objects

3 - Using orthogonalization

Find the intersection of

  • the plane spanned by [1, 0, 0] and [0, 1,−1]
  • the plane spanned by [1, 2,−2] and [0, 1, 1]

The orthogonal complement in <math>\mathbb{R}^3</math> of the first plane is Span {[4,−1, 1]}.

  • Therefore first plane is <math>\{[x, y, z] \in \mathbb{R}^3 : [4,−1, 1] · [x, y, z] = 0\}</math>

The orthogonal complement in <math>\mathbb{R}^3</math> of the second plane is Span {[0, 1, 1]}.

  • Therefore second plane is <math>\{[x, y, z] \in \mathbb{R}^3 : [0, 1, 1] · [x, y, z] = 0\}</math>

The intersection of these two sets is the set: <MATH> \{[x, y, z] \in \mathbb{R}^3 : [4,−1, 1] · [x, y, z] = 0 \text{ and } [0, 1, 1] · [x, y, z] = 0\} </MATH> By Row-Space/Null-Space Duality, a basis for this vector space is a basis for the null space of <math>A = \begin{bmatrix} \begin{array}{rrr} 4 & −1 & 1 \\ \hline \\ 0 & 1 & 1 \\ \end{array} \end{bmatrix} </math>

The null space of A is the orthogonal complement of Span {[4,−1, 1], [0, 1, 1]} in <math>\mathbb{R}^3</math>… which is Span {[1, 2,−2]}

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linear_algebra/find_intersection.txt · Last modified: 2017/11/30 10:43 by gerardnico