Answer to Question #214062 in Analytic Geometry for Dee

Question

(5.1) Find the vector form of the equation of the plane that passes through the point P0 = (1, −2, 3) (5)

and has normal vector ~n =< 3, 1, −1 >.

(5.2) Find an equation for the plane that contains the line x = −1 + 3t, y = 5 + 3t, z = 2 + t and is (6)

parallel to the line of intersection of the planes x −2(y −1) + 3z = −1 and y = −2x −1 = 0.

Answer 1

(5.1) The standard form of the equation of the plane is

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

$$3x+y-z+2=0$$

Parameters: $y=s, z=t$

$$x=-\dfrac{2}{3}-\dfrac{1}{3}s+\dfrac{1}{3}t$$

$$y=0+1s+0t$$

$$z=0+0s+1t$$

The vector form of the equation of the plane is

$$\vec r=\langle\dfrac{2}{3},0,0\rangle+s\langle-\dfrac{1}{3},1,0\rangle+t\langle\dfrac{1}{3},0,1\rangle$$

(5.2)

$$x −2(y −1) + 3z = −1$$ $$y = −2x −1$$

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

$$y=-2x-1$$ $$z=-\dfrac{5}{3}x-\dfrac{5}{3}$$ $$x\in \R$$

The intersection between the planes is the line :

$$\langle0,-1,-\dfrac{5}{3}\rangle+t\langle1, -2,-\dfrac{5}{3}\rangle, t\in \R$$

The the plane contains the line

$$\dfrac{x-0}{1}=\dfrac{y+1}{-2}=\dfrac{z+\dfrac{5}{3}}{-\dfrac{5}{3}}$$

An equation for the plane parallel to the line of intersection of the planes

$$ax+by+cz+d=0,$$

where

$$a-2b-\dfrac{5}{3}c=0$$

The plane contains the line $x = −1 + 3t, y = 5 + 3t, z = 2 + t$

$t=0: Point(-1, 5, 2)$

$t=-2: Point(-7, -1, 0)$

$$-a+5b+2c+d=0$$ $$-7a-b+d=0$$

$$b=-7a+d$$ $$-a-35a+5d+2c+d=0$$

$$b=-7a+d$$ $$c=18a-3d$$

Substitute

$$a+14a-2d-30a+5d=0$$

$$d=5a$$

If $a=1$

$$d=5$$

$$b=-2$$

$$c=3$$

The equation for the plane is

$$x-2y+3z+5=0,$$