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    I've been given the following. The angle ACO is 15 degrees and BCO is 30 degrees.

    Find the lengths of OB and OC, done that  1+\frac{1}{3}(2\sqrt3) and  2+\sqrt3 respectively. Then I'm asked to find the coordinates of vectors AB AC and BC.

    Now I'm not too sure on this part, as I haven't been given any points. All I've managed to do is rewrite it as OB-OA etc.
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    Eh. I'd just find the length of OA and go from there.

    Assuming A, B, and C are on the x, y, z axis respectively, vector OA = (OA,0,0), vector OB = (0,OB,0), and vector OC = (0,0,OC).
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    (Original post by aznkid66)
    Eh. I'd just find the length of OA and go from there.

    Assuming A, B, and C are on the x, y, z axis respectively, vector OA = (OA,0,0), vector OB = (0,OB,0), and vector OC = (0,0,OC).
    So I've got the length of OA OB and OC (OA is 1 shown on the diagram), but then how do I find the coordinates? is AB just (0,1,0)-(1,0,0)?
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    As you said, AB=-OA+OB.

    If OA lies on the ? axis and has a magnitude of 1, then what're the three components of vector OA?
    If OB lies on the ? axis and has a magnitude of 2/rt3+1, then what're the three components of vector OB?

    (I only use ? because your diagram doesn't clearly show exactly which points lies on which axis.)
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    (Original post by aznkid66)
    As you said, AB=-OA+OB.

    If OA lies on the ? axis and has a magnitude of 1, then what're the three components of vector OA?
    If OB lies on the ? axis and has a magnitude of 2/rt3+1, then what're the three components of vector OB?

    (I only use ? because your diagram doesn't clearly show exactly which points lies on which axis.)
    Yeah sorry diagram is rubbish. A lies on the x axis B on the y C on the z.

    so the components of OA are (1,0,0) OB are (1,2/rt3+1,0) OC (0,0,2+root3) ?
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    (Original post by Music99)
    so the components of OA are (1,0,0) OB are (0,2/rt3+1,0) OC (0,0,2+root3) ?
    Fixed that for ya, so yup :P
    So you see how to arrive at the answers now, right?
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    (Original post by aznkid66)
    Fixed that for ya, so yup :P
    So you see how to arrive at the answers now, right?
    :P cheers. Yeah was just being a bit silly, might quote you again in a bit if I get stuck on anymore if you don't mind?
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    (Original post by Music99)
    :P cheers. Yeah was just being a bit silly, might quote you again in a bit if I get stuck on anymore if you don't mind?
    Sure, no problem ^^
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    (Original post by aznkid66)
    Sure, no problem ^^
    Find the area of the triangle ABC I did that got 19.3 (3sf), Then I'm asked to find the coordinates of M, where the normal from the origin meets the plane of the triangle ABC. Is that the point that just passes through C on my diagram?
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    Hm...well, wouldn't it be more inside the triangle rather than on it's vertex? ^^;
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    (Original post by aznkid66)
    Hm...well, wouldn't it be more inside the triangle rather than on it's vertex? ^^;
    I'm not too sure .
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    (Original post by Music99)
    I'm not too sure .
    Well, wouldn't it look like this (imagine the vertex more right-angley please :P)?

    Name:  Untitled.jpg
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    (Original post by aznkid66)
    Well, wouldn't it look like this (imagine the vertex more right-angley please :P)?

    Name:  Untitled.jpg
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    But in the original diagram I've got O is where M is, so I thought it would be where the O is in your diagram?
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    Wait, woah, what? Isn't O the origin? O isn't in the triangle, right?
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    (Original post by aznkid66)
    Wait, woah, what? Isn't O the origin? O isn't in the triangle, right?
    It just says with the point O at the origin.
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    But M is on the plane, in fact it's the foot of the normal from the plane to the origin?
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    (Original post by aznkid66)
    But M is on the plane, in fact it's the foot of the normal from the plane to the origin?
    Yeah, I'm pretty sure that's right.
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    ...Anyway, what have you tried?
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    Derp.

    Do you know how to find the vector representing the shortest distance (the perpendicular to the plane) between a point and a plane?
 
 
 
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