Find the equation of the plane that contains:

Question

Find an equation for the plane containing the lines $$x = 5y = \frac{z + 1}{4}$$ and $$\begin{cases} x = t \\ y = 2t\\ z = 6t − 1 \end{cases}.$$

I know that finding two points will allow me to find the plane, but how do I go about finding those points?

Answer 1

with $t=0$ we get the point $(x,y,z)=(0,0,-1)$ and with $t=-1$ we get $(x,y,z)=(-1,-2,-7)$ for the other line setting $z=3$ and we obtain $x=1,y=\frac{1}{5}$ and you have got three points.

Answer 2

Remember that the line $$\frac{x-x_0}{a} = \frac{y-y_0}{b}=\frac{z-z_0}{c}$$ passes through $(x_0,y_0,z_0)$ with direction $(a,b,c)$, and the same for the line $$\begin{cases} x = x_0+at \\ y = y_0 + bt \\ z = z_0 + ct\end{cases}.$$ To find the plane, you need one point, and a normal vector. To find the normal vector, you can do the cross product between the directions of the two lines, and you can find the intersection by inspection, or by substituition.

Answer 3

The direction vector of the first line is:

$$\textbf{v}_1=\left(1, \frac{1}{5},4\right)$$

The direction vector of the second line is:

$$ \textbf{v}_2 = (1, 2, 6) $$

The cross product of these two vectors is normal to the plane:

$$ \textbf{n} = \textbf{v}_1 \times \textbf{v}_2 $$