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

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The discussion revolves around determining the intersection of a line and a plane, specifically where the line defined by z=<1,2,-3>+t(2,-3,-2) intersects the plane 2x+3y+z=12. The initial answer provided was 3,-1,-5, which was later confirmed to be incorrect due to an error in calculating the parameter t. After substituting the line equations into the plane equation and solving for t, the correct intersection point was found to be -1,5,-1. The conversation emphasizes the importance of careful calculations and sign management in vector problems.
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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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