# Thread: angle between plane and plane

1. ## angle between plane and plane

can someone explain why when z=0 , n =k https://www.flickr.com/photos/123101...3/14096908607/

2. ## Re: angle between plane and plane

Originally Posted by delso
can someone explain why when z=0 , n =k https://www.flickr.com/photos/123101...3/14096908607/
The plane $z=0$ is the $xy-$plane, $\{(x,y,0)\}$.
The vector $\vec{k}=<0,0,1>$ is normal to each of those $<x,y,0>$.

3. ## Re: angle between plane and plane

do u mean in order to get the scalar product =0, you equate (x, y, 0 ) dot (0, 0 , 1) to 0 ? but in order to get 0 , why must the value be 1 ? the unknown value cant be other than 1?

4. ## Re: angle between plane and plane

Originally Posted by delso
do u mean in order to get the scalar product =0, you equate (x, y, 0 ) dot (0, 0 , 1) to 0 ? but in order to get 0 , why must the value be 1 ? the unknown value cant be other than 1?
How much do you know about vector geometry?
Do you know that $\vec{i}=<1,0,0>,~\vec{j}=<0,1,0>,~\&~\vec{k}=<0,0 ,1>,$ is the basics for vector spaces.

$\vec{i}$ is the normal for the $yz$-plane; $\vec{j}$ is the normal for the $xz$-plane; $\vec{k}$ is the normal for the $xy$-plane.

It seems to me that you are very confused about all of this material.
Is this an online course?

5. ## Re: angle between plane and plane

If you are still confused, perhaps this will help. The coefficients for $\displaystyle x$ and $\displaystyle y$ are both zero. So, the plane $\displaystyle z=0$ is the same as

$\displaystyle 0\cdot x + 0\cdot y + 1\cdot z = 0$

So, the normal is:

$\displaystyle \vec{n} = 0\cdot \vec{i} + 0\cdot \vec{j} + 1 \cdot \vec{k}$

You find the normal for that plane just as you would find the normal for any other plane.

6. ## Re: angle between plane and plane

sorry my basic of vector is too weak. i cant really undersatnd while i 'm studying this chapter .