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If $A,B,C,D$ are $4$ coplanar points and $A',B',C',D'$ are their projections on any plane. If $\alpha$ is the angle between plane of $ABCD$ and the plane of projections. Then the ratio of volume of volume of Tetrahedron $AB'C'D'$ to volume of Tetrahedron $A'BCD$ is

what i try

Let $\vec{A'B'}=x,\vec{A'C'}=y,\vec{A'D'}=z$

Then volume of Tetrahedron $\displaystyle V_{AB'C'D'}=\frac{1}{6}[\vec{AB'}\;\vec{AC'}\;\vec{AD'}]$

Let $\vec{AB}=u,\vec{AC}=v\;,\vec{AD}=w$

Then volume of tetrahedron $\displaystyle V_{A'BCD}=\frac{1}{6}[\vec{A'B}\;\vec{A'C}\;\vec{A'D}]$

did not know how to solve it

Help me please

jacky
  • 5,194

2 Answers2

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The area of $\triangle B'C'D'$ is $\cos\alpha$ times the area of $\triangle BCD$. The height of $A'$ above $\triangle BCD$ is $\cos\alpha$ times the height of $A$ above $\triangle B'C'D'$. These two factors cancel, so the two tetrahedra have the same volume.

joriki
  • 238,052
1

Without loss of generality, assume A, B, C and D are in the $xy$-plane. Then, the projections of B, C and D onto an arbitrary place can be represented by the vectors

$$\vec{OB'} = \vec{OB} + b\vec z, \>\>\>\>\>\vec{OC'} = \vec{OC} + c\vec z, \>\>\>\>\>\vec{OD'} = \vec{OD} + d\vec z$$

where $\vec z$ is the unit vector along the $z$-direction. Note that $a$, $b$ and $c$ determine the projection plane and the corresponding A' is the intersection of the vertical line passing A and the plane. Then, $\vec{A'A}=a\vec z$ and the side vectors are,

$$\vec{C'B'} = \vec{OB'} - \vec{OC'} = \vec{CB} + (b-c)\vec z$$ $$\vec{C'D'} = \vec{OD'} - \vec{OC'} = \vec{CD} + (d-c)\vec z$$

As a result, the volume of the Tetrahedron AB'C'D' is

$$V_{AB'C'D'} = \frac16 (\vec{C'B'}\times \vec{C'D'} ) \cdot \vec{A'A}$$

$$ = \frac16 \left\{[\vec{CB} + (b-c)\vec z]\times [\vec{CD} + (d-c)\vec z] \right\} \cdot \vec{A'A}$$

$$ = \frac16 \vec{CB}\times \vec{CD} )\cdot \vec{A'A} + \left[ (b-c)\vec z\times \vec{CD} +(d-c) \vec{CB}\times \vec z +(b-c)(d-c)\vec z\times \vec z\right] \cdot a\vec z$$

Note that $\vec z\times \vec z=0$, $(\vec z\times \vec{CD})\cdot \vec{z}= (\vec{CB}\times \vec z)\cdot \vec{z}=0$. Thus,

$$V_{AB'C'D'} = \frac16 (\vec{C'B'}\times \vec{C'D'} ) \cdot \vec{A'A} =\frac16 (\vec{CB}\times \vec{CD} )\cdot \vec{A'A}=V_{A'BCD}$$

Quanto
  • 97,352