Question

# Find an equation for the plane containing the two (parallel) lines v_1=(0,1,-2)+t(2,3,-1) and v_2=(2,-1,0)+t(2,3,-1).

Vectors
Find an equation for the plane containing the two (parallel) lines
$$v_1=(0,1,-2)+t(2,3,-1)$$
and $$v_2=(2,-1,0)+t(2,3,-1).$$

2021-06-11
The given two lines are,
$$v_1=(0,1,-2)+t(2,3,-1)$$ and
$$v_2=(2,-1,0)+t(2,3,-1)$$
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 $$\vec{n}=\vec{AB}\times\text{direction ratio of the line}$$
$$\vec{n}=(2,-2,2)\times(2,3,-1)$$
$$\vec{n}=\begin{bmatrix}i&j&k\\2&-2&2\\2&3&-1\end{bmatrix}=-4i+6j+10k$$
Thus the equation of the plane can be written as,
$$[(x-0)i+(y-1)j+(z+2)\cdot\vec{n}=0$$
$$-4(x-0)+6(y-1)+10(z+2)=0$$
$$-4x+6y+10z+14=0$$