# Thread: A vertical tower on a horizontal plane- but then it tilts!

1. ## A vertical tower on a horizontal plane- but then it tilts!

Alright I'm a bit stuck here, because I keep going in circles because I feel like I don't have enough information to complete the problem. I know there's got to be a trick somewhere! The question is:

A million years ago, an alien species built a vertical tower on a horizontal plane. When they returned they discovered that the ground had tilted so that measurements of 3 points on the ground gave coordinates of (0, 0, 0), (3, 2, 0), and (0, 3, 1). By what angle does the tower now deviate from the vertical?

Alright, so I drew myself a picture, got the normal for the new plane (it's (2,3,9)) but now I'm stuck. If I could have the normal for the original plane, I could just stick it into the geometric formula for dot products and get the angle between the two...

What am I missing here?!

2. Find the equation of the plane with the given coordinates and compare it to the vertical plane. That is when the tower was vertical.

Or use the horizontal plane when it set on level ground.

Then use $cos{\theta}=\frac{(-n_{1})\cdot (n_{2})}{||n_{1}||\cdot ||n_{2}||}$

where $n_{1}, \;\ n_{2}$ are the normals of said planes.

3. Yeah, that's what I'm talking about when I say using the geometric dot product formula. Except HOW do I compare that to the vertical plane? The equation will just give me 90 degrees, because that's what the normal is- perpendicular to the plane.

I'm sorry it's really hard to get this across, I'm not very good with putting things into words when I'm thinking hard.

4. Hello, robotobunneh!

It's very simple . . . don't kick yourself.

A million years ago, an alien species built a vertical tower on a horizontal plane.
When they returned they discovered that the ground had tilted so that measurements
of 3 points on the ground gave coordinates of $(0, 0, 0),\;(3, 2, 0),\;(0, 3, 1).$
By what angle does the tower now deviate from the vertical?

Alright, so I drew myself a picture, got the normal for the new plane; it's $\langle 2,3,9\rangle$
but now I'm stuck. If I could have the normal for the original plane,
I could just stick it into the formula and get the angle between the two.

The normal of a horizontal plane is straight up.
. . its normal vector is: . $\langle 0,0,1\rangle$.

5. Wow, thank you so much! I was thinking along those lines, but for some reason couldn't get there (I blame midterm stress left over from last week)!

Thanks : )