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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

#./interaufg69_en.tex#The plane

$\displaystyle E: x+y=1 $

dissects the tetrahedron with vertices

$\displaystyle (x, y, z) = (0,0,0), (2,0,0), (0,3,0), (0,0,6) $

into two solid subfigures.


\includegraphics[width=.3\linewidth]{TdM_13_A3_bild}

Find the points of intersection $ A, B, C, D$ on the edges.
Compute the volumes of the tetrahedron and of the solid subfigure containing the origin $ O$.

Hint: The dashed triangle $ \triangle (C, P, Q)$ dissects one of the solid subfigures into a prism and a pyramid having a quadrilateral base.


Answer:

$ A = $ ( , , )    
$ B = $ ( , , )    
$ C = $ ( , , )    
$ D = $ ( , , )    
$ V_{\text{tetrahedron}} = $    
$ V_{\text{subfigure}} = $    
     

(The results should be correct to three decimal places.)

Solution:

We obtain the points of intersection on the edges by inserting the parametric representation of the edge into the equation of the plane.

$ A: (t,0,0) \to E \Rightarrow t = 1,$ i. e.

$\displaystyle A = (1,0,0) $

$ B: (0,t,0) \to E \Rightarrow t = 1,$ i. e.

$\displaystyle B = (0,1,0) $

$ C: (1-t)(2,0,0) + t(0,0,6) \to E \Rightarrow 2-2t = 1,$ i. e. $ t = 1/2$ and

$\displaystyle C = (1,0,3) $

$ D: (1-t)(0,3,0) + t(0,0,6) \to E \Rightarrow 3-3t = 1,$ i. e. $ t = 2/3$ and

$\displaystyle D = (0,1,4) $


The area of the base triangle of the tetrahedron is

$\displaystyle 1/2 \cdot 2 \cdot 3 = 3 . $

Hence, for the volume of the tetrahedron we get

$\displaystyle V_{\text{tetrahedron}} = \dfrac{1}{3} \cdot 3 \cdot \vert OR\vert = \dfrac{1}{3} \cdot 3 \cdot 6 = 6 , $

with $ R = (0,0,6)$ being the vertex of the tetrahedron.




The prism with base $ \triangle (O, A, B)$ and height $ \vert A C\vert = 3$ has the volume

$\displaystyle V_1 = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}. $

The base of the pyramid is a trapezoid with vertices $ D, P, Q, R$.
Its area is

$\displaystyle \dfrac{1}{2}(\vert RQ\vert+\vert PD\vert) \cdot \vert QP\vert = \dfrac{1}{2} (3+1) \cdot 1 = 2 . $

Multiplying the area with one third of the height $ \vert CQ\vert$, we obtain the volume

$\displaystyle V_2 = \dfrac{1}{3} \cdot 1 \cdot 2 = \dfrac{2}{3}. $


Hence, the solid subfigure containing the origin $ O$ has the volume

$\displaystyle V_{\text{subfigure}} = V_1 + V_2 = \dfrac{3}{2}+\dfrac{2}{3} = \dfrac{13}{6} \approx 2.167. $

[problem of the week]