Vector inteception..

[geoff18]

1. The problem statement, all variables and given/known data
where does the line z=<1,2,-3>+t(2,-3,-2) intersect the plane 2x+3y+z=12

i got the answer as 3,-1,-5



2. Relevant equations
i follow the steps listed here:
http://www.physicsforums.com/showthread.php?t=277585


3. The attempt at a solution

i got the answer as 3,-1,-5

can someone please answer if the answer is correct or not, if not please explain what i did wrong.
many thanks

[Chewy0087]

Not quite....what value did you get for t?

It appears that you've used a value of t being 1, unfortunately it isn't. Your working should look something like this;

x for the line = (1+2t)
y for the line = (2-3t)
z for the line = (-3-2t)

Then inserting for t in the plane as in replacing 2x+3y+z=12 with the values above and solving for t, having done that putting it back into the line equation to get;

(1,2,-3) + t(2, -3,-2)

[geoff18]

hi thx for the prompt reply

these are my working out:
x=1+2t
y=2-3t
z=-3-2t

then i sub it in

2(1+2t)+3(2-3t)+(-3-2t)-12=0
i got t=-7/7=/-1

[geoff18]

If a≠0, t=-b/a

So the point where the line intersects the plane is:

LaTeX Code: M(x_1 - \\frac{b}{a}a_1 , y_1 - \\frac{b}{a}a_2 , z_1 - \\frac{b}{a}a_3)

so i switch:
1-(-1)(2),(2)-(-1)(-3),(-3)-(-1)(-2)
which i got
3,-1,-5

[geoff18]

so is the answer -1,5,-5?
anyone?

[Chewy0087]

so is the answer -1,5,-5?
anyone?

Right apart from the last one, be careful with signs.

[geoff18]

so its' -1,5,-1?
thx chewy

[Chewy0087]

That's right, no problem.