## 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 \( A_{1} A_{2} A_{3} \).

\( A_{1}(2 ; -4 ; -3) \)

\( A_{2}(5 ; -6 ; 0) \)

\( A_{3}(-1 ; 3 ; -3) \)

\( A_{4}(-10 ; -8 ; 7) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{5-2 ;-6-(-4) ; 0-(-3)\}=\{3 ;-2 ; 3\} \\
& A_{1} A_{3}=\{-1-2 ; 3-(-4) ;-3-(-3)\}=\{-3 ; 7 ; 0\} \\
& A_{1} \overrightarrow{A_{4}}=\{-10-2 ;-8-(-4) ; 7-(-3)\}=\{-12 ;-4 ; 10\}
\end{aligned}
$$

According to the geometric meaning of the mixed product, we have:

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

We compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
3 & -2 & 3 \\
-3 & 7 & 0 \\
-12 & -4 & 10
\end{array}\right|= \\
& =3 \cdot\left|\begin{array}{cc}
7 & 0 \\
-4 & 10
\end{array}\right|-(-2) \cdot\left|\begin{array}{cc}
-3 & 0 \\
-12 & 10
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
-3 & 7 \\
-12 & -4
\end{array}\right|= \\
& =3 \cdot 70--2 \cdot(-30)+3 \cdot 96=210-60+288=438
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|438|=73
$$

Since

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

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

We compute the vector product:

$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
3 & -2 & 3 \\
-3 & 7 & 0
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-2 & 3 \\
7 & 0
\end{array}\right|-j\left|\begin{array}{cc}
3 & 3 \\
-3 & 0
\end{array}\right|+k \cdot\left|\begin{array}{cc}
3 & -2 \\
-3 & 7
\end{array}\right|= \\
& =i \cdot(-21)-j \cdot 9+k \cdot 15=\{-21 ;-9 ; 15\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-21)^{2}+(-9)^{2}+15^{2}}=\frac{\sqrt{747}}{2}
$$

Then

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 73}{\frac{\sqrt{747}}{2}}=\frac{438}{\sqrt{747}}$

Volume of the tetrahedron: 73

Height: $\frac{438}{\sqrt{747}}$

## Problem Kuznetsov Analytic Geometry 7-30