Question

What is the relationship between the volumes of the tetrahedron defined by the vectors

$\left[\begin{array}{c}{a}_1\\ a_2\\ 1\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{c}{b}_1\\ b_2\\ 1\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{c}{c}_1\\ c_2\\ 1\end{array}\right]$

and the area of the triangle with vertices

$\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]\mathbf{?}$

See Exercises 4 and 5. Explain this relationship geometrically. Hint: Consider the top face of the tetrahedron.

Short answer

Therefore, therelationship between volume of tetrahedron and area of triangle is given by,

$V_2=3V_1$

Step-by-step solution

Definition.

Area of the triangle:

Basically, it is equal to half of the Area of the parallelogram.

The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle.

$A=1/2\times \mathrm{b}\times \mathrm{h}$

Area of tetrahedrons:

It is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners.

It is also known as a triangular pyramid.

General formula of the volume of the tetrahedrons is

$V=1/3(\mathrm{area}\mathrm{of}\mathrm{base})(\mathrm{perpendicular}\mathrm{height})$

To find the relationship between volume of tetrahedron and area of triangle.

From what we know from ex. 5, the area of a tetrahedron defined by

$\left[\begin{array}{c}{a}_1\\ a_2\\ 1\end{array}\right],\left[\begin{array}{c}{b}_1\\ b_2\\ 1\end{array}\right],\left[\begin{array}{c}{c}_1\\ c_2\\ 1\end{array}\right]$

Is

role="math" localid="1660721138671" $$\begin{gathered} V_1=\frac{1}{6}\left|det\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ 1& 1& 1\end{array}\right]\right| \\ {V}_1=\frac{1}{6}\left|(b_1{c}_2-c_1{b}_2)-(a_1{c}_2-c_1{a}_2)+(a_1{b}_2-b_1{a}_2)\right| \end{gathered}$$

On the other hand, the area of a triangle with vertices

$\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right],\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right],\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$

Is

$$\begin{gathered} V_2=\frac{1}{2}\left|det\left[\begin{array}{cc}{b}_1-a_1& c_1-a_1\\ b_2-a_2& c_2-a_2\end{array}\right]\right| \\ {V}_2=\frac{1}{2}\left|b_1{c}_2-a_2{b}_1-a_1{c}_2-b_2{c}_1+a_1{b}_2+a_2{c}_1\right| \end{gathered}$$

Therefore,

$V_2=3V_1$