1. ## Angles between planes

I need to find the angle between the planes x-z=1 and 2y+z=1

So the normal vectors for the planes would be
<1,0,-1) and <0,2,1) Correct?

And the magnitudes of those vectors would be
(2)^(1/2) and (5)^(1/2) Correct?

Thus Cos(Theta)= -1/((2^.5)*(5^.5))
Correct?

Leaving me with an angle of 1.8925 Radians?

Is this all correct?

2. ## Re: Angles between planes

Looks good to me, though I've been having trouble since my hari-kari.

3. ## Re: Angles between planes

Well, this is the wrong answer...unless the teacher is wrong. So I'm trying to figure out what's wrong here.

4. ## Re: Angles between planes

Originally Posted by Bracketology
Well, this is the wrong answer...unless the teacher is wrong. So I'm trying to figure out what's wrong here.
Is your teacher's answer $1.249~?$
That is the acute angle between the plane,

5. ## Re: Angles between planes

Yes. How do I find this angle?

6. ## Re: Angles between planes

Originally Posted by Bracketology
Yes. How do I find this angle?
It usual in such a question to find the acute angle or maybe right angle.
$\Theta = \arccos \left( {\frac{{\left| {u \cdot v} \right|}}{{\left\| u \right\| \cdot \left\| v \right\|}}} \right)~.$

7. ## Re: Angles between planes

Originally Posted by Bracketology
Yes. How do I find this angle?
Pi - 1.89 (!)

8. ## Re: Angles between planes

When two planes (or two lines) intersect they make four angles but only two distinct angles- the "opposite" (vertical) angles are congruent. The two distinct angles between two planes add to $\pi$ radians. $\pi- 1.8925=1.2491$ radians.

As to why that formula did not immediately give you the correct answer, note that <-1, 0, 1> is also normal to the first plane and if you had used that, you would have got $cos(\theta)= \frac{1}{\sqrt{2}\sqrt{5}}$. I.e., to get the smaller angle, use the absolute value.

9. ## Re: Angles between planes

Thanks for all the help guys!