Parametric line in 3D: point and plane intersection

(a) Parametric equation

$$\mathbf P(u)=\mathbf P_1+u(\mathbf P_2-\mathbf P_1),\qquad 0\le u\le1$$

$\mathbf P_2-\mathbf P_1=(4,\,-4,\,6)$, so

$$x=1+4u,\qquad y=2-4u,\qquad z=3+6u$$

(b) Length and direction cosines

$$L=\sqrt{4^2+(-4)^2+6^2}=\sqrt{68}=8.246$$ $$l=\frac{4}{8.246}=0.485,\quad m=\frac{-4}{8.246}=-0.485,\quad n=\frac{6}{8.246}=0.728$$

Check: $l^2+m^2+n^2=0.235+0.235+0.529=1.0$.

(c) Point at u = 0.25

$$x=1+1=2,\quad y=2-1=1,\quad z=3+1.5=4.5$$

Point $(2,\ 1,\ 4.5)$.

(d) Intersection with z = 6

$3+6u=6 \Rightarrow u=0.5$, which lies in $0\le u\le1$, so the segment does cut the plane.

$$x=1+4(0.5)=3,\quad y=2-4(0.5)=0$$

Answer: L = 8.246; direction cosines (0.485, -0.485, 0.728); P(0.25) = (2, 1, 4.5); intersection with z = 6 at (3, 0, 6).