# Vectors - Finding a point of intersection

1. Apr 11, 2012

### dimens

1. The problem statement, all variables and given/known data
The line L is parallel to the vector 3i - 2j -2k and passes through the point P (1,0,-1/2)

find the point of intersection Q of the line L with the plane ∏

x+y+z=2

3. The attempt at a solution
I'm completely stumped with this, don't know where to start... I thought maybe

(x,y,z) = (1,0,1/2) + 2(3,-2,-2)
= (7,-4,7/8)

but it's wrong.. Where do I start?

2. Apr 11, 2012

### tiny-tim

welcome to pf!

hi dimens! welcome to pf!

hint: what is a parametric equation for L ?

3. Apr 11, 2012

### dimens

Thanks for the welcome, I'm probably gonna be on here a lot to help me get through university. Lol.

Parametric equations:

x=1+3t
y=-2t
z=1/2-2t

4. Apr 11, 2012

### tiny-tim

ok, now that has to coincide with the equation x + y + z = 2 …

sooo … t = … ?

(btw, is that 1/2 or -1/2?)

5. Apr 11, 2012

### dimens

x+y+z = 2...

so...

1 + 3t -2t +1/2 - 2t = 2
-t=0.5
t=0.5

plug it back into the equation right?

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

(5/2,-1,-1/2)

... Answers wrong though, it's meant to be...

(-7/2,3,-5/2)

Last edited: Apr 11, 2012
6. Apr 11, 2012

### dimens

x=1+3t
y=-2t
z=-1/2-2t

x+y+z = 2

....

1+3t -2t -1/2 -2t = 2
t= -3/2

(x,y,z) = (1,0,-1/2) + -3/2(3,-2,-2)
(x,y,z) = (1,0,-1/2) + (-9/2,3,3)

= (-7/2,3,5/2)

I got it :) Cheers tiny tim. I think I learn a lot better when I'm discussing things and typing it out on here.

7. Apr 11, 2012

### dimens

continuing the question:

find the angle the line L makes with the plane ∏ at the intersection point Q.

a.b = |a||b| cos∅

(3,-2,-2) (-7/2,3,5/2)

... a.b = -21.5
|a| = sqrt(17)
|b| = sqrt(27.5)

∅ = arccos (-21.5/sqrt(467.5))

... answer is wrong. Am I missing something?

8. Apr 11, 2012

### tiny-tim

that's the wrong line, it has nothing to do with the plane ∏