Paralellepiped n angles n gaahhh n HELP ME

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  • #1
brandy
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i don't know what this angle on my paralellpiped is. basicallt just a trapazoid extruded on an angle. but also half of the front face is pushed back. i need to no what angle it is pushed back.
dimensions are:
10 by 8 with 120 60 degress (front n back face)
10 by 6 with 120 n 60 degress (sides)
6 by 8 with 135 n 45 degrees (top n bottom)

specificly i want to find the plotting points on a 3 dimensional plane (x y z).
where the front faces has the lower left most point as the origin. i need to find out what the points would be for the back face.
 

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  • #2
brandy
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please help. i still can't figure it out. i tried using 30 degress (90-60) (60 being the angle on the face not the actual plane so it failed) i neetheth help!
 
  • #3
tiny-tim
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Welcome to PF!

specificly i want to find the plotting points on a 3 dimensional plane (x y z)

Hi brandy! Welcome to PF! :smile:

(erm … what's a "3 dimensional plane"? :redface:)

Hint: the end of the 10-long side starting at the origin is at (x,y), where the inner products (x,y).U and (x,y).V = 10cos60º, where U and V are the unit vectors of the other two sides at the origin. :smile:
 
  • #4
brandy
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oops i meant on a three dimensional...i forget the word. just something with height depth and width. and i already knew that stuff about unit vectors. besides that's on one of the faces of the shape. I am trying to figure out the height of the whole thing. so the length between the top and bottom faces (6cm 45degress 8 cm 135 degress faces) one way i could figure this out is if i knew what the angle was between the face (10 60 8 120 faces (front and back)) and the z axis along the x axis. and that's my question. what's the angle between the front face of the shape and the line along the x and up the z axises.
 
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  • #5
tiny-tim
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Hi brandy! :smile:

(oops … I meant (x,y,z).U and (x,y,z).V = 10cos60º, of course :redface:)
im trying to figure out the height of the whole thing …

The height is just the z coordinate of that endpoint (x,y,z).

And the angle of the front face will have tangent z/y. :smile:
…i forget the word …

I'd say "in three-dimensional Cartesian coordinates" (as opposed to spherical coordinates, for example). :smile:
 
  • #6
brandy
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gahh. i don't know the coodinantes this is just a shape we had to make so all that i know is the angles ON THE FACES and the legnths of the size. the front face is on angle about 17 degrees parralell to the y and z axis. therefore the legnth along that face will NOT be the hieght. can u tell me how i could work out what the height is? apart from just measuring it?
 
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  • #7
tiny-tim
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Hi brandy! :smile:
gahh. i don't know the coodinates

Yes, but you will know them if you use the unit vector method!

Try it and see! :smile:
 
  • #8
brandy
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ther are to many unkowns to work it out. (for my tiny brains anyhow). my teacher said it was had something to do with taking the angle at the origin of the front face and minusing the angle on the the top face. so...
 
  • #9
tiny-tim
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ok … let's call three of the sides OU OV and OW, with OU along the axis, and OV in the x-y-plane, O = (0,0,0), and U = (8,0,0).

Then what is V?

And then what is W (using (x,y,z).U and (x,y,z).V = 10cos60º)? :smile:
 
  • #10
brandy
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read the words i am writing. i do not know what the z coodinant is!. the angle is NOT 60 degrees. the angle is UNKNOWN. U is 8 ? ?. A PARALELLAPIPED HAS NO PERPENDICULAR FACES. think about it. that means that their are no 90degree angles anywhere. NONE OF THE FACES ARE ALONG ANY AXIS'S except for the base and top face which is along the x-y plane. the front face is on an incline but you can't see it front on. if u turned it side on you will see it is on an incline. as u would i u had turned on its front.
 
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  • #11
tiny-tim
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… one step at a time …

i do not know what the z coodinant is!. …

brandy … calm down … be in your happy place! … :smile:

let's go one step at a time …

First step: if O is (0,0,0), and U is (8,0,0), and V is in the x-y-plane, then what is V? :smile:
 

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