Finding the angle between surfaces

l46kok

1. The problem statement, all variables and given/known data


*Length of each side of the square is 3.
*Length of DP = Length of BQ = Length of GR = 1

Calculate [tex]\theta[/tex], the angle that is made between the surface of PQR and the surface of CGHD

2. Relevant equations

No relevant equation was given for me, but I'm going to throw out some that may help.

*Law of Cosines
*Dot Product

3. The attempt at a solution

I do not have a single clue of what to do. Can anyone start me off somehow, and tell me what I should be initially looking for in order to find the angle between the surfaces?

robphy

The angle between the surfaces is equal to the angle between the normal-vectors to these surfaces... which you can determine using the dot-product.
You have enough information to determine each normal-vector.

l46kok

Wait, but I thought normal vector was found using the cross product, not the dot product?

robphy

That's one way to do it (and that's the best way for your problem).
Now, once you found those normal vectors using the cross-product,
now use the dot product to find the angle between those [normal] vectors.

l46kok

Well I think I can find the normal vector of the flat surface, but how would you find the normal vector of the triangular surface?

robphy

Can you find two linearly-independent [i.e. two non-parallel] vectors on that triangular surface?

l46kok

Pq = (3,-1,0) Pr = (3,0,-3)

l46kok

where do I go from here

robphy

Use the cross-product, as you said in #3.

l46kok

? I don't understand how that would help me, as that would produce a vector that is normal to the two vectors given above, NOT the normal vector that is perpendicular to the surface.

I'll do it anyways.

(3,9,3)

HallsofIvy

If two vectors lie IN a surface, then the normal to the surface must be normal to both. In fact, unless the two vectors lie along the same line (in which case their cross product would be 0) their cross product IS normal to the surface.

l46kok

Ok, if that is the case, let me try to solve this

Dc = (3,0,0) Dh = (0,0,-3)

Dc X Dh = (0,9,0) = A

Pq X Pr = (3,9,3) = B

[tex]A * B = |A||B|\cos(\theta)[/tex]

[tex]\cos^{-1}\frac{A * B}{|A||B|} = \theta[/tex]

[tex]\theta = \cos^{-1}\frac{81}{|9||9.798|}[/tex]

[tex]\theta = 23.284288 degrees[/tex]

...And that is an incorrect answer, where did I go wrong?

HallsofIvy

The FIRST thing you had better do is set up a coordinate system so that vectors from one point to another make sense! Set up a coordinate system with origin at F, x-axis along FG, y-axis along FE, z-axis along FB. Then the vector perpendicular to CGHD is <1,0, 0> and the vector perpendicular to plane PQR is the cross product of QP= <1,0,1>-<2,1,1>= <-1, -1, 0> and QR= <1,0,1>- <2,0,0>= <-1,0,1>. That cross product is <-1,1,-1>. The angle is given by <1, 0, 0>.<-1, 1, -1>= |<1,0,0>||<1,1,1>|cos([itex]\theta[/itex] or -1= (1)([itex]\sqrt{3}[/itex])cos([itex]\theta[/itex])which gives cos([itex]\theta)= -1/\sqrt{3}[/itex]. What is [itex]\theta[/itex]?

l46kok

[tex]\theta = 125.26439 degrees[/tex]

And according to the answer, it is still wrong.

The answer is supposed to be

[tex]\cos(\theta) = \frac{3\sqrt{11}}{11}[/tex]

l46kok

I can't find anything you did wrong either. Would the answer be wrong?

msslowlearner

the normals as you calculated were right .. maybe the simplification of the cos theta part may have gone wrong ...

i proceeded as hallsofvy said in #13, but i got cos theta = -3/sqrt 11

Char. Limit

Remember that the usual range of the arccosine function is [0,pi]. So you'll need to find a POSITIVE value of theta. Also, remember that cos(x)=cos(-x).