## Task Condition

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

$A_{1}(1 ; 2 ; 0)$

$A_{2}(3 ; 0 ;-3)$

$A_{3}(5 ; 2 ; 6)$

$A_{4}(8 ; 4 ;-9)$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{3-1 ; 0-2 ;-3-0\}=\{2 ;-2 ;-3\} \\
& \overrightarrow{A_{1} A_{3}}=\{5-1 ; 2-2 ; 6-0\}=\{4 ; 0 ; 6\} \\
& \overrightarrow{A_{1} A_{4}}=\{8-1 ; 4-2 ;-9-0\}=\{7 ; 2 ;-9\}
\end{aligned}
$$

According to the geometric meaning of the mixed 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 compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
2 & -2 & -3 \\
4 & 0 & 6 \\
7 & 2 & -9
\end{array}\right|= \\
& =2 \cdot\left|\begin{array}{cc}
0 & 6 \\
2 & -9
\end{array}\right|-(-2) \cdot\left|\begin{array}{cc}
4 & 6 \\
7 & -9
\end{array}\right|+(-3) \cdot\left|\begin{array}{cc}
4 & 0 \\
7 & 2
\end{array}\right|= \\
& =2 \cdot(-12)+2 \cdot(-78)-3 \cdot 8=-24-156-24=-204
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-204|=34
$$

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 compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
2 & -2 & -3 \\
4 & 0 & 6
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-2 & -3 \\
0 & 6
\end{array}\right|-j\left|\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right|+k \cdot\left|\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right|= \\
& =i \cdot(-12)-j \cdot 24+k \cdot 8=\{-12 ;-24 ; 8\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-12)^{2}+(-24)^{2}+8^{2}}=\frac{1}{2} \cdot \sqrt{784}=\frac{28}{2}=14
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 34}{14}=\frac{51}{7}=7 \frac{2}{7}
$$

Volume of the tetrahedron: 34

Height: $7 \frac{2}{7}$

## Problem Kuznetsov Analytic Geometry $7-11$