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Math Help - tetrahedron 2

  1. #1
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    tetrahedron 2

    Given the tetrahedron points A(1,2,1), B(2,-1,1), C(0,-1,-1) and D(3,1,0). Calculate the measure of height low of vertex D to the plan of the face ABC.




    Answer:
    \frac{8}{\sqrt{19}}
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  2. #2
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    Hello, Apprentice123!

    Given points: A(1,2,1),\:B(2,-1,1),\:C(0,-1,-1),\:D(3,1,0)
    Calculate the distance from vertex D to the plane of face ABC.

    Answer: . \frac{8}{\sqrt{19}}
    First, find the equation of the plane . . .

    We have: . \begin{array}{ccccc}\vec u & = & \overrightarrow{AB} &=& \langle 1,\text{-}3,0\rangle \\ \vec v &=& \overrightarrow{BC} &=& \langle \text{-}2,0,\text{-}2\rangle \end{array}

    The normal vector of the plane: . \vec n \:=\:\vec u \times \vec v \:=\:\left[\begin{array}{ccc}i & j & k \\ 1 &\text{-}3 & 0 \\ \text{-}2 &0 & \text{-}2 \end{array}\right] \;=\;6i + 2j - 6k
    Hence: . \vec n \:=\:\langle 3,1,\text{-}3\rangle


    The plane is: . 3(x-1) + 1(y-2) - 3(z-1) \:=\:0 \quad\Rightarrow\quad 3x + y - 3z - 2 \:=\:0


    The distance from point (x_1,y_1,z_1) to plane ax + by + cz + d \:=\:0

    . . is given by: . D \;=\;\frac{|ax_1 +by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}


    So we have: . D \;=\;\frac{|3(3) + 1(1) - 3(0)|}{\sqrt{3^2 + 1^2 + (\text{-}3)^2}} \;=\;\boxed{\frac{8}{\sqrt{19}}}

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  3. #3
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    thank you
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