Find parametric equations for the line of intersection of the planes 3x-2y+z=1,

CheemnCatelvew 2021-10-23 Answered

Find parametric equations for the line of intersection of the planes \(3x-2y+z=1\), \(2x+y-3z=3\)

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Expert Answer

Laith Petty
Answered 2021-10-24 Author has 14940 answers
\(\displaystyle{3}{x}-{2}{y}+{z}={1}\)
\(\displaystyle{2}{x}+{y}-{3}{z}={3}\)
FInd a point of intersection. Set z=0, solve for x and y.
\(\displaystyle{3}{x}-{2}{y}+{0}={1}\to{x}={\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}\)
\(\displaystyle{2}{x}+{y}-{0}={3}\)
\(\displaystyle{2}\cdot{\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}+{y}={3}\)
\(\displaystyle{2}{\left({2}{y}+{1}\right)}+{3}{y}={9}\)
\(\displaystyle{7}{y}+{2}={9}\)
\(\displaystyle{7}{y}={7}\)
\(\displaystyle{y}={1}\)
\(\displaystyle{x}={\frac{{{1}}}{{{3}}}}{\left({2}{\left({1}\right)}+{1}\right)}={1}\)
Point 1: \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}={\left({1},{1},{0}\right)}\)
We can use the cross product to find a direction vector for the line.
The cross product of the normals to the planes results in a vector that is orthogonal to both, and would be parallel to the line of intersection.
\(\displaystyle{<}{a},{b},{c}{>}\times{<}{d},{e},{f}\ge{<}{b}{f}-{c}{e},{c}{d}-{a}{f},{a}{e}-{b}{d}{>}\)
\(\displaystyle{<}{3},-{2},{1}{>}\times{<}{2},{1},-{3}\ge{<}-{2}{\left(-{3}\right)}-{1}{\left({1}\right)},{1}{\left({2}\right)}-{3}{\left(-{3}\right)},{3}{\left({1}\right)}-{\left(-{2}\right)}{\left({2}\right)}{>}\)
\(\displaystyle={<}{5},{11},{7}{>}\)
Plug the point into \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}\) and the vector found into \(\displaystyle{<}{a},{b},{c}{>}\)
\(\displaystyle{x}={x}_{{0}}+{a}{t}\)
\(\displaystyle{y}={y}_{{0}}+{b}{t}\)
\(\displaystyle{z}={z}_{{0}}+{c}{t}\)
\(\displaystyle{x}={1}+{\left({5}\right)}{t}\)
\(\displaystyle{y}={1}+{\left({11}\right)}{t}\)
\(\displaystyle{z}={0}+{\left({7}\right)}{t}\)
\(\displaystyle{x}={1}+{5}{t}\)
\(\displaystyle{y}={1}+{11}{t}\)
\(\displaystyle{z}={7}{t}\)
Result: \(\displaystyle{x}={1}+{5}{t}\ {y}={1}+{11}{t}\ {z}={7}{t}\)
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