How to calculate the intersection of two planes ? These are

tornesasln 2021-11-20 Answered
How to calculate the intersection of two planes ?
These are the planes and the result is gonna be a line in \(\displaystyle{\mathbb{{{R}}}}^{{3}}\):
\(\displaystyle{x}+{2}{y}+{z}-{1}={0}\)
\(\displaystyle{2}{x}+{3}{y}-{2}{z}+{2}={0}\)

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Expert Answer

barcelodurazo0q
Answered 2021-11-21 Author has 483 answers
You need to solve the two equations
\(\displaystyle{x}+{2}{y}+{z}-{1}={0}\)
\(\displaystyle{2}{x}+{3}{y}-{2}{z}+{2}={0}\)
Notice that, these are two equations in three variables, so you have a free variable say \(\displaystyle{z}={t}\), then we have
\(\displaystyle{x}+{2}{y}={1}-{t}\)
\(\displaystyle{2}{x}+{3}{y}={2}{t}-{2}\)
Solving the last system gives
\(\displaystyle{\left\lbrace{x}=-{7}+{7}{t},{y}={4}-{4}{t}\right\rbrace}\)
Then the parametrized equation of the line is given by
\(\displaystyle{\left({x},{y},{z}\right)}={\left(-{7}+{7}{t},{4}-{4}{t},{t}\right)}={\left(-{7},{4},{0}\right)}+{\left({7},-{4},{1}\right)}{t}\)
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Roger Noah
Answered 2021-11-22 Author has 1113 answers
The first plane has normal vector \[(\begin{array}{c}1\\2\\1\end{array})\] and the second has normal vector \[(\begin{array}{c}2\\3\\-2\end{array})\], so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors \(\displaystyle{v}_{{1}},\ {v}_{{2}}\) is \(\displaystyle{v}_{{1}}\times{v}_{{2}}\), so the direction vector of the line of intersection is
\[(\begin{array}{c}1\\2\\1\end{array})\times(\begin{array}{c}2\\3\\-2\end{array})=(\begin{array}{c}-7\\4\\-1\end{array})\]
Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try y=0(since it passes through the x-z plane) and solve the resulting system of linear equations:
\(\displaystyle{x}+{z}-{1}={0}\)
\(\displaystyle{2}{x}-{2}{z}+{2}={0}\)
giving \(\displaystyle{x}={0},{z}={1}\), thus the line of intersection is \[\{(\begin{array}{c}-7t\\4t\\1-t\end{array}):t\in\mathbb{R}\}\]
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