# Find the intersection of a particle and a plane in R3

## 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.

## The Attempt at a Solution

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Mark44
Mentor
Are there any relevant equations (e.g., equation of a line in space, parametric equation of a plane in space, angle between two vectors)?

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?

Mark44
Mentor
I'm trying to get you to fill in the information you neglected to enter in the first post.

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.

Mark44
Mentor
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.

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

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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Mark44
Mentor
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.

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)?

Mark44
Mentor
Check and see. Does this point satisfy your equations for the line and the plane? If so, you're golden.