Question

1.   Find the vector form of the equation of the plane that passes through the point P0 = (1, −2, 3) and has normal vector ~n =< 3, 1, −1 >.

2.   Find an equation for the plane that contains the line x = −1 + 3t, y = 5 + 3t, z = 2 + t and is parallel to the line of intersection of the planes x −2(y −1) + 3z = −1 and y = −2x −1 = 0.

3.   Find the point of intersection between the lines: < 3, −1, 2 > + t < 1, 1, −1 > and <−8, 2, 0 > + t < −3, 2, −7 >.

4.   Show that the lines x + 1 = 3t, y = 1, z + 5 = 2t for t ∈ R and x + 2 = s, y − 3 = −5s, z + 4 = −2s for t ∈ R intersect, and find the point of intersection.

5.   Find the point of intersection between the planes: −5x + y −2z = 3 and 2x −3y + 5z = −7.

Answer 1

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

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

2.

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

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

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

The intersection between the planes is the line :

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

The line

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

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 that 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

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

3.

$$3+s=-8-3t$$ $$-1+s=2+2t$$ $$2-s=0-7t$$ $$s=-3t-11$$ $$4=-5t-10$$ $$1=-5t+2$$

$$s=-3t-11$$ $$3=-12$$ $$1=-5t+2$$

No solution.

There is no point of intersection.

4.

$$3t-1=s-2$$ $$1=-5s+3$$ $$2t-5=-2s-4$$

$$s=3t+1$$ $$5s=2$$ $$2t=-2s+1$$

$$t=-\dfrac{1}{5}$$ $$s=\dfrac{2}{5}$$ $$t=\dfrac{1}{10}$$

No solution.

There is no point of intersection.

5.

$$-5x+y-2z=3$$ $$2x-3y+5z=-7$$

$$-13x-z=2$$ $$2x-3y+5z=-7$$

$$x=-\dfrac{2}{13}-\dfrac{1}{13}z$$

$$y=\dfrac{2}{3}x+\dfrac{5}{3}z+\dfrac{7}{3}$$

$$x=-\dfrac{2}{13}-\dfrac{1}{13}z$$

$$y=\dfrac{29}{13}+\dfrac{21}{13}z$$

$$z\in\R$$

The intersection between the planes is the line :<3, - 1,2> +<1, 1, - 1>

$$\langle-\dfrac{2}{13}, \dfrac{29}{13}, 0\rangle+t\langle-\dfrac{1}{13}, \dfrac{1}{13},1\rangle, t\in \R$$