intersection between lines in R3?

Question

L1 : (x+3)/1 = (y-7)/4 = (z-2)/1

L2 : a line goes through (0,2,1) and (-4,4,1)

i found the parametric equations

L1: x = -3 - t, y = 7 + 4t, z = 2 + t
L2: x = -4s, y = 2 + 2s, z = 1

i set L1 = L2 but i can't find any s and t values that satisfy.

I used t = -1 initially and plugged them into L1's parametric equation and got the answer in the back (-2,3,1), but if i solve for s which I got, was s = 2, i get (-8,6,1).

HELP!!!

typo:
L1 should be is (x+3)/-1

Answer 1

Find the intersection of the two lines.

L1 : (x + 3)/-1 = (y - 7)/4 = (z - 2)/1
L2 : a line goes through (0,2,1) and (-4,4,1)

Let's rewrite the equations of both lines in parametric form.

L1:
s = (x+3)/-1 = (y-7)/4 = (z-2)/1

x = -3 - s
y = 7 + 4s
z = 2 + s

L2:
The directional vector v, of the line is:
<-4-0, 4-2, 1-1> = <-4, 2, 0>

Any non-zero multiple of the directional vector is also a directional vector. Divide by -2.

v = <2, -1, 0>

With a point and the directional vector we can write the equation of the line. Let's use the first point.

L2 = <0, 2, 1> + t<2, -1, 0>

x = 0 + 2t
y = 2 - t
z = 1
_______

Set the equations of the line equal to each other seperately by variable and solve for s and t to find the point of intersection.

x = -3 - s = 0 + 2t
y = 7 + 4s = 2 - t
z = 2 + s = 1

Solving we get:

z = 2 + s = 1
s = -1

x = -3 - s = 0 + 2t
-3 - (-1) = 2t
-2 = 2t
t = -1

Plug the solution for s and t into the equation for y to see if it is consistent. If it is we have a point of intersection.

y = 7 + 4s = 2 - t
7 + 4(-1) = 2 - (-1)
3 = 3

The point of intersection is:

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

The point of intersection is P(-2, 3, 1).

Answer 2

It is L1: x = -3 + t, y = 7 + 4t, z = 2 + t
and not x =-3-t

ok, you set for example z=2+t=1 so t=-1

If t=-1, now you set x=-3-t=-4s

so x = -2 = -4s ,
s=1/2

And indeed y=7 + 4t = 2 + 2s, when s=1/2 and t=-1
Hence the lines intersect in (-2,3,1).