Answer to Question #204308 in Analytic Geometry for Frank

Question

The planes pie1 and pie2 have cartesian equations x-y+z+1=0 and 2x+y-z-1=0 respectively.

(a) Show that these two planes are not parallel.

(b) Find a parametric equation of their line of intersection

(c)Give a cartesian equation of plane pie3 passing through point A(1,-1,-1) and perpendicular to pie1 and pie2.

Answer 1

a) two planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ are parallel if $\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2} = \dfrac{C_1}{C_2} .$

In our case $\dfrac{A_1}{A_2} = \dfrac{1}{2}, \; \dfrac{B_1}{B_2} = \dfrac{-1}{1}, \; \dfrac{C_1}{C_2} = \dfrac{1}{-1}$ , these numbers are not the same, so the planes are not parallel.

b) the coordinates of points on the line of intersection satisfy the equations

$\begin{cases} x-y+z+1 = 0, \\ 2x+y-z-1 = 0 \end{cases}$

$\begin{cases} x-y+z=-1, \\ 2x+y-z=1 \end{cases}$

The equivalent matrices:

$\begin{bmatrix} 1 & -1 & 1 & -1 \\ 2 & 1 & -1 & 1 \end{bmatrix}$ , $\begin{bmatrix} 1 & -1 & 1 & -1 \\ 0 & -\dfrac32 & \dfrac32 & -\dfrac32 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -\dfrac32 & \dfrac32 & -\dfrac32 \end{bmatrix},$

$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 &1 \end{bmatrix},$

$\begin{cases} x = 0 + 0t, \\ y = 1 + t,\\ 0 = 0 + t. \end{cases}$

c) Let the equation of the third plane be $A_3x+B_3y+C_3z + D_3 = 0.$

It is perpendicular to pi_1 and pi_2, so

$1\cdot A_3 - 1\cdot B_3 + 1\cdot C_3 = 0, \\ 2\cdot A_3 + 1\cdot B_3 - 1\cdot C_3 = 0$ , therefore $A_3 = 0,$ $B_3 = C_3.$ Let $B_3 = 1.$

A(1,-1,-1) belongs to plane, so $A_3\cdot 1 + B_3\cdot (-1) + C_3\cdot(-1) + D_3 = 0,$

$0 - 1 - 1 + D_3 = 0, D_3 = 2.$

So the equation of the plane is $y+z+2 = 0.$