Solving Vector Interception Homework: 3,-1,-5

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Homework Help Overview

The problem involves determining the intersection of a line defined by the equation z=<1,2,-3>+t(2,-3,-2) with a plane described by the equation 2x+3y+z=12. The original poster claims to have found the intersection point as (3,-1,-5).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the value of t used in the calculations and question the correctness of the intersection point provided by the original poster. They explore substituting the line's parametric equations into the plane's equation to solve for t.

Discussion Status

There is an ongoing examination of the calculations, with some participants providing guidance on how to substitute values into the plane's equation. Multiple interpretations of the results are being explored, particularly regarding the signs and values derived from the calculations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance provided. There is a focus on verifying the correctness of the original poster's answer and clarifying the steps taken to arrive at that conclusion.

geoff18
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Homework Statement


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



Homework Equations


i follow the steps listed here:
https://www.physicsforums.com/showthread.php?t=277585


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
 
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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)
 
hi thanks 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
 
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
 
so is the answer -1,5,-5?
anyone?
 
geoff18 said:
so is the answer -1,5,-5?
anyone?

Right apart from the last one, be careful with signs.
 
so its' -1,5,-1?
thx chewy
 
That's right, no problem.
 

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