Answer to Question #213618 in Analytic Geometry for Tshego

Question

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

b.) 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.

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

D.)let L be the line given by <3, - 1,2> +t<1,1-1>, for t € R.

1.) show that the above line L lies on the plane - 2x + 3y - 4z +1 =0

2.)Find an equation for the plane through the point P =(3, - 2,4)that is perpendicular to the line <-8, 2, 0> +t<-3,2,-7>

Answer 1

a)

$$3+s=-8-3t$$ $$-1+s=2+2t$$ $$2-s=0-7t$$

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

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

No solution.

There is no point of intersection.

b)

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

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

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

No solution.

There is no point of intersection.

с)

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

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

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

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

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

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

$$z\in\R$$

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

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

d)

1)

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

$$-6-2t-3+3t-8+4t+1$$

$$=5t-16\not=0\ for\ t=0$$

The line does not lie on the plane.

2) Line

$$\dfrac{x+8}{-3}=\dfrac{y-2}{2}=\dfrac{z-0}{-7}$$

Point $P=(3, -2, 4)$

The equation for the plane through the point P that is perpendicular to the line is

$$-3(x-3)+2(y-(-2))-7(z-4)=0$$

$$-3x+9+2y+4-7z+28=0$$

$$3x-2y+7z-41=0$$