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

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

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Laith Petty
$$\displaystyle{3}{x}-{2}{y}+{z}={1}$$
$$\displaystyle{2}{x}+{y}-{3}{z}={3}$$
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: $$\displaystyle{x}={1}+{5}{t}\ {y}={1}+{11}{t}\ {z}={7}{t}$$