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Math Help - Finding Normal Angle of 2d Triangle in 3 dimensions

  1. #1
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    Finding Normal Angle of 2d Triangle in 3 dimensions

    Hello, so I've been trying to find out how to find the normal angle of a triangle when only knowing the X,Y and Z coordinates of each corner. The triangle will always be a right triangle. If you took a square and cut it from bottom left to top right, it would be one of the two triangles made from that. Anyways, I know the X Y and Z of each corner and the hyp will always be the square root of two (sides are 1). I've never done angles in 3 dimensions, but I think I will need to find a normal angle for XY and Z seperately. Even if that isn't necessary, I need two separate angles. Thanks for any help! Also the angle would be on the face of the triangle facing upwards.
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    Quote Originally Posted by Sidelines View Post
    Hello, so I've been trying to find out how to find the normal angle of a triangle when only knowing the X,Y and Z coordinates of each corner. The triangle will always be a right triangle. If you took a square and cut it from bottom left to top right, it would be one of the two triangles made from that. Anyways, I know the X Y and Z of each corner and the hyp will always be the square root of two (sides are 1). I've never done angles in 3 dimensions, but I think I will need to find a normal angle for XY and Z seperately. Even if that isn't necessary, I need two separate angles. Thanks for any help! Also the angle would be on the face of the triangle facing upwards.
    So you have three points

    P_1=(x_1,y_1,z_1)
    P_2=(x_2,y_2,z_2)
    P_3=(x_3,y_3,z_3)

    So use the distance formula to find the distance between
    P_1P_2 \quad P_1P_3 \quad P_2P_3

    d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

    Do this for the others as well and you will know all of the side length of the triangle. Remember the largest side is always the hypotenuse
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    Thanks, but I'm looking for the normal angle off the triangle. not the lengths. Unless that's how you get the normal angle, through taking sin and cos? but I think it's a little more complicated? but I could be wrong...
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    Quote Originally Posted by Sidelines View Post
    Thanks, but I'm looking for the normal angle off the triangle. not the lengths. Unless that's how you get the normal angle, through taking sin and cos? but I think it's a little more complicated? but I could be wrong...
    That is what I was thinking.
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    Well the lengths of the sides will always be the same, but the triangle could be rotated on any axis to any degree. And I have no clue where to begin in finding the normal angle that something would reflect off of the face of this triangle with haha...
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    I think I really misunderstood what you were attempting to do. This link may be helpful.
    math - Find the normal angle of the face of a triangle in 3D, given the co-ordinated of its verticies - Stack Overflow
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    Oh yep! That's almost exactly my problem! Thanks, but if you could, could you explain what it means by taking the cross product of the two vectors? I'm not exactly sure how to go about a vector with 3 components.
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    Quote Originally Posted by Sidelines View Post
    Oh yep! That's almost exactly my problem! Thanks, but if you could, could you explain what it means by taking the cross product of the two vectors? I'm not exactly sure how to go about a vector with 3 components.
    Given two vectors we can take their cross product as follows

    \mathbf{v}=<v_1,v_2,v_3>
    \mathbf{w}=<w_1,w_2,w_3>

    \mathbf{v} \times \mathbf{w}=\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}=<v_2w_3-v_3w_2,v_3w_1-v_1w_3,v_1w_2-v_2w_1>
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    Ah, I see. Well I really hope this works haha... but the math seems right! Thanks for everything!
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