# Find an equation for the plane containing the two (parallel) lines v_1

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

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Alix Ortiz

The given two lines are,
$$\displaystyle{v}_{{1}}={\left({0},{1},-{2}\right)}+{t}{\left({2},{3},-{1}\right)}$$ and
$$\displaystyle{v}_{{2}}={\left({2},-{1},{0}\right)}+{t}{\left({2},{3},-{1}\right)}$$
Let A=(0,1,-2)
B=(2,-1,0)
Thus the plane containing these two lines will contain A and B and hence the vector AB=(2,-2,2)
Normal to the plane shall be the cross product of AB and the direction ratio of the lines.
Thus $$\displaystyle\vec{{{n}}}=\vec{{{A}{B}}}\times\text{direction ratio of the line}$$
$$\displaystyle\vec{{{n}}}={\left({2},-{2},{2}\right)}\times{\left({2},{3},-{1}\right)}$$
$$\vec{n}=\begin{bmatrix}i&j&k\\2&-2&2\\2&3&-1\end{bmatrix}=-4i+6j+10k$$
Thus the equation of the plane can be written as,
$$\displaystyle{\left[{\left({x}-{0}\right)}{i}+{\left({y}-{1}\right)}{j}+{\left({z}+{2}\right)}\cdot\vec{{{n}}}={0}\right.}$$
$$\displaystyle-{4}{\left({x}-{0}\right)}+{6}{\left({y}-{1}\right)}+{10}{\left({z}+{2}\right)}={0}$$
$$\displaystyle-{4}{x}+{6}{y}+{10}{z}+{14}={0}$$