The angle of intersection between two planes in R3

[Erenjaeger]

What is the angle of intersection between the two planes in ℝ³ with general equations
x-y=5 and y-z=7?

I know that the angle between then is equal to cos^-1 (u⋅v/||u|| ||v||) but I am stuck on the general equations given, how can I solve when given these ?

 

[Mark44]

What is the angle of intersection between the two planes in ℝ³ with general equations
x-y=5 and y-z=7?

I know that the angle between then is equal to cos^-1 (u⋅v/||u|| ||v||) but I am stuck on the general equations given, how can I solve when given these ?

Find a normal to each plane, and then use the dot product to find the cosine of the angle between the two normals. That will be the same as the angle between the two planes.

 

[Erenjaeger]

Find a normal to each plane, and then use the dot product to find the cosine of the angle between the two normals. That will be the same as the angle between the two planes.

how would i find a normal to each plane?

 

[Erenjaeger]

how would i find a normal to each plane?

im confused with those general equations given in the problem

 

[Erenjaeger]

Find a normal to each plane, and then use the dot product to find the cosine of the angle between the two normals. That will be the same as the angle between the two planes.

In my course book, we are given an example where we have two points in ℝ³ which are (2,-1,3) and is parallel to (3,2,-2) so the parametric equations are
x=2+3t
y=-1+2t
z=3-2t
where t∈ℝ
solving for t gives

t=x-2/3, t=y+1/2, t=3-z/2

and this yields two equations
x-2/3 = y+1/2 and y+1/2 = 3-z/2 which I can understand because they're all the same value which is 't' so they are all equal to one another right?

the part where I am lost is where it says rearranging these two equations gives some general equations for the two planes...
2x-3y=7 and y+z=2
how do they rearrange to get those equations ?

 

[chiro]

Hey Erenjaeger.

This is done by solving a linear system (i.e. a "matrix") and finding what occurs when you reduce it down to "row echelon form".

Have you ever done linear algebra before?

 

[Erenjaeger]

Hey Erenjaeger.

This is done by solving a linear system (i.e. a "matrix") and finding what occurs when you reduce it down to "row echelon form".

Have you ever done linear algebra before?

yeah

 

[chiro]

It's basically just an application of standard matrix methods.

 

[Erenjaeger]

It's basically just an application of standard matrix methods.

okay, ill read through that section in my course book, thanks

 

[Mark44]

how would i find a normal to each plane?

Given a plane in R³, Ax + By + Cz = D, a normal to this plane is the vector <A, B, C>. My suggestion, which doesn't involve matrices, is a lot simpler than the one chiro gave.

Find a normal to each plane: $\vec{n_1} = <A_1, B_1, C_1>$ and $\vec{n_2} = <A_2, B_2, C_2>$ and use the fact that $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}|\cos(\theta)$, where $\theta$ is the angle between the two vectors.