What's the intersection point of two lines given in canonical form?

Question

I hope someone can help me with this:

What's the intersection point of two lines given in canonical form?

$ D1 : \frac{x}{2} = \frac{y}{-3} = \frac{z}{1} $

$ D2 : \frac{x+1}{3} = \frac{y+5}{2} = \frac{z}{1} $

Answer 1

write your System in the form $$x=2t$$ $$y=-3t$$ $$z=t$$ and $$x=3s-1$$ $$y=2s-5$$ $$z=s$$ from the last equation we get $$t=s$$ so we get $$s=t=1$$ and we get the searched Point as $$P(2;-3;1)$$

Answer 2

$\frac{x}{2} = \frac{x+1}{3} \rightarrow x = 2, \frac{y}{-3} = \frac{y+5}{2} \rightarrow y = -3, z = 1$. Thus, such point is $(2, -3, 1)$

Answer 3

Considering that the lines you give have an intersection point, all you have to do is: from D1 you get: 3x + 2y = 0. From D2 you get: 2x - 3y = 13. Find x,y by solving the system and then take for example the equation x/2 = z(of line D1) to find z by replacing x from the solution.(you should get the same z from D2,too).