Intersection between a line and a plane.

Question

A line can either lie on a plane, lie parallel to it or intersect it. Determine, if there is one, the point of intersection between the line given by the equation

$$\displaystyle\frac{x−5}{2} =\displaystyle\frac{y−1}{-1} = \displaystyle\frac{z−15}{4}$$
and the plane given by the equation $$(x, y, z) = (-2, -7, 5) + s(2, 6, 3) + t(1, 4, -1)$$

So, what I have to do, is determine if the line and the plane either intersect or are parallel? What equation applies in this problem?

Answer 1

A very basic way could be to find $x$, $y$ and $z$ from plane's equation and then put them into line's equation. This would lead us to see if the achieved system of equations is consistent or not: $$x=-2+2s+t, y=-7+6s+4t, z=5+3s-t$$

Answer 2

By inspection we can find that the line is parallel to the vector $(2,-1,4)$ and that the point $(5,1,15)$ lies on the line. In homogeneous coordinates, the line is thus the join of $\mathbf p=[2:-1:4:0]$ and $\mathbf q=[5:1:15:1]$. This line can be represented by its Plücker matrix $\mathbf p\mathbf q^T-\mathbf q\mathbf p^T$.

A normal to the plane is $(2,6,3)\times(1,4,-1)=(-18,5,2)$ and $(-18,5,2)\cdot(-2,-7,5)=11$, so the plane can be represented in homogeneous coordinates as $\mathbf\pi=[-18:5:2:-11]$. The intersection of the line and plane is $$L\mathbf\pi=\mathbf p\mathbf q^T\mathbf\pi-\mathbf q\mathbf p^T\mathbf\pi=-66\mathbf p+33\mathbf q=[33:99:231:33]$$ or in (inhomogeneous) Cartesian coordinates, $(1,3,7)$.