## problem statement

Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4}$ and its height dropped from vertex $A_{4}$ to the face $A_{1} A_{2} A_{3}$.

$A_{1}(0 ;-1 ;-1)$

$A_{2}(-2 ; 3 ; 5)$

$A_{3}(1 ;-5 ;-9)$

$A_{4}(-1 ;-6 ; 3)$
## Solution

From vertex $A_{1}$, we will draw vectors:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-0 ; 3-(-1) ; 5-(-1)\}=\{-2 ; 4 ; 6\} \\
& \overrightarrow{A_{1} A_{3}}=\{1-0 ;-5-(-1) ;-9-(-1)\}=\{1 ;-4 ;-8\} \\
& \overrightarrow{A_{1} A_{4}}=\{-1-0 ;-6-(-1) ; 3-(-1)\}=\{-1 ;-5 ; 4\}
\end{aligned}
$$

According to the geometric meaning of the scalar triple product, we have:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)\right|
$$

We will compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-2 & 4 & 6 \\
1 & -4 & -8 \\
-1 & -5 & 4
\end{array}\right|= \\
& =-2 \cdot\left|\begin{array}{cc}
-4 & -8 \\
-5 & 4
\end{array}\right|-4 \cdot\left|\begin{array}{cc}
1 & -8 \\
-1 & 4
\end{array}\right|+6 \cdot\left|\begin{array}{cc}
1 & -4 \\
-1 & -5
\end{array}\right|= \\
& =-2 \cdot(-56)-4 \cdot(-4)+6 \cdot(-9)=112-(-16)+(-54)=74
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|74|=\frac{37}{3}=12 \frac{1}{3}
$$

Since

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{3} \cdot S_{A_{1} A_{2} A_{3}} \cdot h \Rightarrow h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}
$$

According to the geometric meaning of the vector product:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot\left|\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right|
$$

We will compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-2 & 4 & 6 \\
1 & -4 & -8
\end{array}\right|=i \cdot\left|\begin{array}{cc}
4 & 6 \\
-4 & -8
\end{array}\right|-j\left|\begin{array}{cc}
-2 & 6 \\
1 & -8
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-2 & 4 \\
1 & -4
\end{array}\right|= \\
& =i \cdot(-8)-j \cdot 10+k \cdot 4=\{-8 ;-10 ; 4\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-8)^{2}+(-10)^{2}+4^{2}}=\frac{1}{2} \cdot \sqrt{180}=3 \sqrt{5}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{37}{3}}{3 \sqrt{5}}=\frac{37}{3 \sqrt{5}}
$$

Volume of the tetrahedron: $12 \frac{1}{3}$

Height: $\frac{37}{3 \sqrt{5}}$

## Problem Kuznetsov Analytic Geometry 7-6