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Find the intersection of a particle and a plane in R3

  • Thread starter tiki84626
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  • #1
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Homework Statement



The starting position of a particle in R3 is (1,1,1) and it's traveling with constant velocity (2,-1,1). Where does it hit the plane {(x,y,z)|x - 2y + z = 4}. And find the angle between the path of the particle and the plane.

Homework Equations




The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
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Are there any relevant equations (e.g., equation of a line in space, parametric equation of a plane in space, angle between two vectors)?

What attempt have you made?
 
  • #3
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yeah i am having trouble figuring out how to find the equation for the path of the particle. From there I think it should be pretty easy, just setting them equal and finding their intersection. and then for the angle between them i can just use the dot product, but with what vectors?
 
  • #4
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I'm trying to get you to fill in the information you neglected to enter in the first post.
 
  • #5
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one can express a line by a pair of linear equations

ax+by+cz+d=0 and a'x+b'y+c'z+d'=0

such that bc'-cb', ca'-ac', and ab'-ba' are not all zero. The line thus defined is parallel to the vector

(bc'-cb', ca'-ac', ab'-ba').

Line through (x0,y0,z0) parallel to the vector (a,b,c):

(x-x0)/a = (y-y0)/b = (z-z0)/c

Now, the angle between two vectors can be obtained by using the dot product:

angle = arccos ((a . b)/|a||b|)



as for my attempt, i am having trouble just getting off the ground.
 
  • #6
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Well, that's a start. One that I had in mind was how do you write the equation of a line in space, give a point P(x0, y0, z0) on the line and a vector v = <A, B, C> with the same direction as the line?

This is directly relevant to your problem, whereas what you wrote about the intersection of two planes isn't.
 
  • #7
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So, the equation for a line in space is: r = r0 +tv = (x0,y0,z0) + t<a,b,c>

but once i have that, how do i solve for their intersection? and the angle between them?

For their intersection, do i just solve the system of equations:

x = x0 + ta
y = y0 + tb
z = z0 + tc
x - 2y + z = 4

but what about t?


and for the the angle between them, can i just use the vector form of the line, and a vector parallel to the plane and then use the dot product?
 
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  • #8
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5,318
Now this is something you should have included in the Relevant equations part in your first post.

Put the data of your problem into the equation r = (x0,y0,z0) + t<a,b,c>

Every point on your plane satisfies
x = 2y -z + 4
y = y
z = z

If the line and the plane intersect, their x coordinates, y coordinates, and z coordinates have to be equal.
 
  • #9
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Ok, so i get the equation for the line:

(x,y,z) = (1 + 2t, 1 - t, 1 + t)

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

So, using the equations for our plane:

x = 2y - z + 4
y = y
z = z

we get that:

y = 1 - t
z = 1 + t
x = 1 + 2t = 2(1 - t) - (1 + t) + 4
t = 4/5

So, is their intersection point (13/5, 1/5, 9/5)?
 
  • #10
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5,318
Check and see. Does this point satisfy your equations for the line and the plane? If so, you're golden.
 

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