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?

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

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

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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.

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Is your teacher's answer
That is the acute angle between the plane,

Yes. How do I find this angle?

Quote:

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Originally Posted by Bracketology

Yes. How do I find this angle?

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It usual in such a question to find the acute angle or maybe right angle.

Quote:

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Originally Posted by Bracketology

Yes. How do I find this angle?

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Pi - 1.89 (!)

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 radians. 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 . I.e., to get the smaller angle, use the absolute value.

Thanks for all the help guys!