Question

Given the two planes x-y+2z-1=0 and 3x+2y-6z+4=0. Find a parametric equation for the intersection. (4)

Answer 1

$$x-y+2z-1=0$$ $$3x+2y-6z+4=0$$

Set $z=0,$ solve for $x$ and $y$

$$x-y-1=0$$ $$3x+2y+4=0$$

$$y=x-1$$ $$3x+2(x-1)+4=0$$

$$x=-0.4$$ $$y=-1.4$$

Point $(-0.4, -1.4, 0)$

$$\vec{n_1}\times\vec{n_2}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1& -1 & 2 \\ 3 & 2 & -6 \end{vmatrix}$$

$$= \vec{i}\begin{vmatrix} -1 & 2 \\ 2 & -6 \end{vmatrix}- \vec{j}\begin{vmatrix} 1 & 2 \\ 3 & -6 \end{vmatrix}+ \vec{k}\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}$$

$$=2 \vec{i}+12 \vec{j}+5 \vec{k}$$

Parametric equation of a line is:

$$x=-0.4+2t,\\ y=-1.4+12t, \\z=5t$$