Intersection of standard line and plane

Question

To find the intersection for $$L:\frac{x-2}3=\frac{y-1}4=\frac{z+1}1$$ and a plane $$3x-y+z=0$$ So actually what are the step by step instructions to solve their intersection?

Answer 1

Equate coordinate functions of the line to a parameter $t$: $$\frac{x-2}3=\frac{y-1}4=\frac{z+1}1=t$$ Then $x=3t+2,y=4t+1,z=t-1$. Substitute these into the plane equation: $$3(3t+2)-(4t+1)+(t-1)=0$$ $$6t+4=0\qquad t=-\frac23$$ Then the intersection coordinates fall easily: $x=0,y=-\frac53,z=-\frac53$.

Answer 2

with $$x=3t+2,y=4t+1,z=t-1$$ we get$$3(3t+2)-4t-1+t-1=0$$ solve this for $t$

Answer 3

If you don’t feel like solving any equations, you can compute this intersection directly by using the Plücker matrix of the line.

By inspection, we know that the point $(2,1,-1)$ lies on the line and that the line is parallel to the vector $(3,4,1)$. Expressed in homogeneous coordinates, the line is thus the join of $\mathbf p=[2:1:-1:1]$ and $\mathbf q=[3:4:1:0]$, which can be represented by the Plücker matrix $L=\mathbf p\mathbf q^T-\mathbf q\mathbf p^T$. The plane is represented in homogeneous coordinates by $\mathbf\pi=[3:-1:1:0]$ and so the intersection of the plane and line is $$L\mathbf\pi=\mathbf p\mathbf q^T\mathbf\pi-\mathbf q\mathbf p^T\mathbf\pi = 6\mathbf p-4\mathbf q=[0:-10:-10:6],$$ which in (inhomogeneous) Cartesian coordinates is the point $\left(0,-\frac53,-\frac53\right)$.