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Reputation: 11

Line of Intersection between 2 Planes

I watched and tried to understand bunch of sites and videos about this, and I came into a weird conclusions and some questions. I need some help to explain which one of the method is right, or even both of them are right (but I got different result from each methods).

I'm sorry that I'm bad at explaining things, the first method is solve the equations normally. But, here the link for the video I tried to learn from https://www.youtube.com/watch?v=o7CfCDkRwfY

Second method is to do cross product for the direction and find the point by set one of the variables as 0. https://www.youtube.com/watch?v=jozabh0lFmo

I tried for this example

x+2y+z−1=0

2x+3y−2z+2=0

and turned out for different answers. Is both of the method are correct, or which one? Thank you.

Upvotes: 0

Views: 514

Answers (1)

duffymo
duffymo

Reputation: 308733

You have two equations with three unknowns.

You can eliminate one variable and solve for a relationship between the remaining two. Let's eliminate z.

Multiply the first equation by 2:

2x + 4y + 2z = 2

Add this to the second equation:

4x + 7y = 0

You can solve for y as a function of x:

y = -4x/7

Substitute this back into the first equation:

x  - 8x/7 + z = 1

Simplify by combining the first and second terms:

-x/7 + z = 1

Solve for z:

z = 1 + x/7

Now you have an equation for the line in 3D space.

-inf <= x <= +inf
y = -4x/7
z = 1 + x/7

Both your equations are satisfied by these two points. Since two points are enough to define a line in Euclidean space I'd say I've got the correct answer.

This line goes through the point (0, 0, 1). It also goes through (7, -4, 2)

Here's a parametric representation of that line for -inf <= t <= +inf:

(x, y, z) = (0, 0, 1) + t*(7, -4, 1)

You can see that when t = 0 (x, y, z) = (0, 0, 1) (first point above) and when t = 1 (x, y, z) = (7, -4, 2) (second point above).

I didn't look at either of your videos. This is how I'd solve it.

This is high school algebra.

Upvotes: 1

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