Answer to Question #214061 in Analytic Geometry for Dee

Question

(3.1)Find the point of intersection between the lines: < 3, −1, 2 > + t < 1, 1, −1 > and

< −8, 2, 0 > + t < −3, 2, −7 >.

(3.2) 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.

(3.3) Find the point of intersection between the planes: −5x + y −2z = 3 and 2x −3y + 5z = −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 :

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