Linear Algebra - Find intersection of geometric objects

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 $\mathbb{R}^3$ of the first plane is Span {[4,−1, 1]}.

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

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

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

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

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