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[Ecrofirt]

If anyone can help. i have no fucking clue how to do this:

Find poinf of intersection for the line

(x+1)/2=((t+3)/3=z/-1

and the plane

3x - 2y + 4z +1 = 0

The answer is (1,0,-1) but I have no idea how the fuck to get it.

actually, (1,0,-1) happens if you make y=0 and then substitute that t value into the x and z equations.

What do I do with that to get the answer?

 

[bud]

you're avatar is just wrong...

 

[Ecrofirt]

that doesn't help

 

[Ecrofirt]

GAF has let me down

 

[goodcow]

GAF has let me down

BURNNNNNNNNNNNNNNNNNNNNNNNN

 

[Shompola]

your line equation looks weird.. can you recheck it.. atleast a paranteses is wrong. Ill get back to you when you have verified it.

 

[YakiSOBA]

2 equations, 3 unknowns... isolate one (z for example), equate them z=z.
2 equations, 2 unknowns... go from there, get numbers for x,y,z = (answer,answer,answer)

 

[Shompola]

I would personally rewrite the line equation in parametric form, i.e. x=someting, y=something, z=something and then just insert x,y,z in the planes equation and solve t. Then insert t in the original line equation and get the insersection point.

 

[Shompola]

(x+1)/2=(t+3)/3=z/-1

we rewrite above equation parmatetric form:

(x+1)/2=((t+3)/3=z/-1
(x+1)*3/2 = t+3 = -3*z
t = (x+1) * 3/2 -3 = -3*z -3

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

Our line is (1,0,-1) + t(2/3,0,-1/3) in parametric form

insert it in the plane equation:

3x - 2y + 4z +1 = 0


3 * (1+2/3*t) - 2 * (0) + 4 * (-1-1/3*t) + 1 = 0
3+2t-4-4/3*t+1 = 0
10/3*t = 0
t = 0

insert t = 0 in the line equation and we get (1,0,-1)

 

[Ecrofirt]

No worries guys, I've already taken the test.

My line equation was supposed to have a y in it, not a t.


Shompola: Thanks. That's pretty much what I had to do, and I figured it out mere minutes before I took the test.