# Intersection between a line and a plane.

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?

• Write the plane equation without parameters, i.e. $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Then use the line equation to solve for two of your variables in terms of the other and plug that in to the equation for your plane. Mar 19, 2015 at 14:23
• Can you calculate a vector that is parallel with line? Can you calculate vector perpendicular to the plane?
– zoli
Mar 19, 2015 at 14:25
• Hmmm, When you say write the plane equation without parameters, you mean: . a(2−5)+b(-1−1)+c(15−4)=0 ? Mar 19, 2015 at 14:41

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: $$x=-2+2s+t, y=-7+6s+4t, z=5+3s-t$$
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, $(1,3,7)$.