# Find an equation for the plane containing the two (parallel)

Find an equation for the plane containing the two (parallel) lines
$$\displaystyle{v}_{{1}}={\left({0},{1},−{2}\right)}+{t}{\left({2},{3},−{1}\right)}$$
$$\displaystyle{v}_{{2}}={\left({2},−{1},{0}\right)}+{t}{\left({2},{3},−{1}\right)}$$

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Joseph Lewis

Given:
$$\displaystyle{v}_{{1}}={\left({0},{1},−{2}\right)}+{t}{\left({2},{3},−{1}\right)}$$
$$\displaystyle{v}_{{2}}={\left({2},−{1},{0}\right)}+{t}{\left({2},{3},−{1}\right)}$$
Let $$\displaystyle{A}={\left({0},{1},-{2}\right)}$$, $$\displaystyle{B}={\left({2},-{1},{0}\right)}$$
Thus, the plane containing these two lines will contain A and B. Hence, the vector $$\displaystyle{A}{B}={\left({2},-{2},{2}\right)}$$
Normal to the plave shall be the cross product of AB and the direction ratio of the lines.
Thus, $$\displaystyle\vec{{{n}}}=\vec{{{A}{B}}}\times{d}{i}{r}{e}{c}{t}{i}{o}{n}\ {r}{a}{t}{i}{o}\ {o}{f}\ {t}{h}{e}\ {l}in{e}$$
$$\displaystyle\vec{{{n}}}={\left({2},-{2},{2}\right)}\times{\left({2},{3},-{1}\right)}$$
$\vec{n}=\begin{bmatrix}i & j & k \\2 & -2 & 2 \\2 & 3 & -1 \end{bmatrix}=-4i+6j+10k$
Thus, the equation of the plane can be written as
$$\displaystyle{\left({x}-{o}\right)}{i}+{\left({y}-{1}\right)}{j}+{\left({z}+{2}\right)}\times\vec{{{n}}}={0}$$
$$\displaystyle-{4}{\left({x}-{0}\right)}+{6}{\left({y}-{1}\right)}+{10}{\left({z}+{2}\right)}={0}$$
$$\displaystyle-{4}{x}+{6}{y}+{10}{z}+{14}={0}$$