Re: Angles between planes

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

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.

Re: Angles between planes

Quote:

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 $\displaystyle 1.249~?$

That is the acute angle between the plane,

Re: Angles between planes

Yes. How do I find this angle?

Re: Angles between planes

Quote:

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.

$\displaystyle \Theta = \arccos \left( {\frac{{\left| {u \cdot v} \right|}}{{\left\| u \right\| \cdot \left\| v \right\|}}} \right)~.$

Re: Angles between planes

Quote:

Originally Posted by

**Bracketology** Yes. How do I find this angle?

Pi - 1.89 (!)

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 $\displaystyle \pi$ radians. $\displaystyle \pi- 1.8925=1.2491$ 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 $\displaystyle cos(\theta)= \frac{1}{\sqrt{2}\sqrt{5}}$. I.e., to get the smaller angle, use the absolute value.

Re: Angles between planes

Thanks for all the help guys!