Say I have a parametric line such as
$$x=4+t$$ $$y=-5-t$$ $$z=4+2t$$
How would I go about finding two planes that intersect that line? So far I have separated the line into a point and its direction
$$P(4,-5,4), v=<1,-1,2>$$
I am not sure on how to proceed from here. Please help?
On
Working in homogeneous coordinates, the line can be expressed as the set of linear combinations $\lambda\mathbf p+\mu\mathbf q$ of the points $\mathbf p=[4:-5:4:1]$ and $\mathbf q=[1:-1:2:0]$, which can be read directly from the parametric equations of the line. (The two points are the homogeneous counterparts of a fixed point on the line and its direction vector.) In matrix form, this can be written $$\begin{bmatrix}4&-5&4&1\\1&-1&2&0\end{bmatrix}^T\begin{bmatrix}\lambda\\\mu\end{bmatrix}.$$ Every plane $\mathbf\pi$ that includes this line includes the point $\mathbf p$ and $\mathbf q$, and the latter occurs iff $\mathbf p^T\mathbf\pi=\mathbf q^T\mathbf\pi=0$. In other words, the set of planes whose intersection is the line is the null space of the above matrix. Row-reduction gives $[6:4:-1:0]$ and $[1:1:0:1]$ as a basis for this null space, hence the equations of two planes whose intersection is the line are $$6x+4y-z=0\\x+y+1=0.$$
To define a plane you need a normal vector and a point on the plane.
$\mathbf n = (1,1,0), p=(4,-5,4)$ will do.
$n\cdot(x,y,z) = n\cdot p$ will define one plane.
$x + y = -1$
$\mathbf n = (0,2,1), p=(4,-5,4)$ can be used for the second
$2y + z = -6$
Alternative.
you can find a non-parallel vector.
$(x,y,z) = (4,-5,4) + t(1,-1,2) + s(1,1,1)$ and either leave it in vector form, or convert it to a parametric form.