# Euclidean geometry-2 planes

1. Jun 9, 2010

### irycio

1. The problem statement, all variables and given/known data
Three-dimensional, euclidean space. We've got 2 non-parallel planes:
$$\vec{OX} \cdot \vec{b_1}=\mu_1$$ and $$\vec{OX} \cdot \vec{b_2}=\mu_2$$. Find all the points Y such that Y lies on the first plane and Y+$$\vec{a}$$ lies on the 2nd one. What did you get?

2. Relevant equations
Come in (3.)

3. The attempt at a solution
$$\vec{OY} \cdot \vec{b_1}=\mu_1$$ AND $$\vec{OY} \cdot \vec{b_2} + \vec{a} \cdot \vec{b_2}=\mu_2$$.
Simple manipulation and we get:
$$\vec{OY} \cdot (\vec{b_1} - \vec{b_2} )= \mu_1 - \mu_2 +\vec{a} \cdot \vec{b_2}$$ (substracted 2nd equation from the 1st one).
And this is where I get stuck, I have no idea how to have $$\vec{OY}$$ on one side and the rest on the other.

E: I'm not even sure that what I've got so far is correct-I believe that those points should all lie on one line, and I've got a plane again. But what I believe in geometry is not necessarily true :)

Last edited: Jun 9, 2010
2. Jun 9, 2010

### lanedance

first consider all the points given by Y + a, this will be a plane parallel to the first offset by the vector a.

The points you want will be the intersection of the offset plane with the 2nd plane

3. Jun 9, 2010

### irycio

I still don't catch it. I mean, I understand and see what you wrote, but have no idea, how to proceed with the exercise. If only it was possible to get a plane equation in the form of OX=a+td1+sd2...

4. Jun 9, 2010

### lanedance

ok, i agree the equation of teh plane is takes a little getting used to

To simplify notation lets call the plane X and any vector form the origin to a point on the plane x = OX, then we have
$$x \cdot b_1=\mu_1$$

now say $|b_1|^2 = \beta_1^2$

take consider the vector $b_1' = \frac{\mu_1}{\beta_1^2} b_1$

clearly that vector is on X, as
$$b_1' \cdot b_1 = \frac{\mu_1}{\beta_1^2} b_1 \cdot b_1 = \mu_1$$

now say you have another vector x on the plane, and consider $x' = x-b_1'$, then:

$$x' \cdot b_1 = (x-b_1') \cdot b_1 = (x \cdot b_1)- (b_1' \cdot b_1) =0$$

so now you know b_1 is normal to the plane X, and b_1' is a vector on X, which should allow you to manipulate the equation of the plane