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

\( A_{1}(14 ; 4 ; 5) \)

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

\( A_{3}(-2 ;-6 ;-3) \)

\( A_{4}(-2 ; 2 ;-1) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-5-14 ;-3-4 ; 2-5\}=\{-19 ;-7 ;-3\} \\
& A_{1} A_{3}=\{-2-14 ;-6-4 ;-3-5\}=\{-16 ;-10 ;-8\} \\
& A_{1} A_{4}=\{-2-14 ; 2-4 ;-1-5\}=\{-16 ;-2 ;-6\}
\end{aligned}
$$

In accordance with the geometric meaning of the scalar triple product, we have:

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-19 & -7 & -3 \\
-16 & -10 & -8 \\
-16 & -2 & -6
\end{array}\right|= \\
& =-19 \cdot\left|\begin{array}{cc}
-10 & -8 \\
-2 & -6
\end{array}\right|-(-7) \cdot\left|\begin{array}{cc}
-16 & -8 \\
-16 & -6
\end{array}\right|+(-3) \cdot\left|\begin{array}{cc}
-16 & -10 \\
-16 & -2
\end{array}\right|= \\
& =-19 \cdot 44+7 \cdot(-32)-3 \cdot(-128)=-836-224+384=-676
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-676|=\frac{338}{3}=112 \frac{2}{3}
$$

Since

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

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

We compute the cross product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-19 & -7 & -3 \\
-16 & -10 & -8
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-7 & -3 \\
-10 & -8
\end{array}\right|-j\left|\begin{array}{cc}
-19 & -3 \\
-16 & -8
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-19 & -7 \\
-16 & -10
\end{array}\right|= \\
& =i \cdot 26-j \cdot 104+k \cdot 78=\{26 ;-104 ; 78\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{26^{2}+(-104)^{2}+78^{2}}=\frac{1}{2} \cdot \sqrt{17576}=\frac{26 \sqrt{26}}{2}=13 \sqrt{26}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{338}{3}}{13 \sqrt{26}}=\frac{26}{\sqrt{26}}=\sqrt{26}$

Volume of the tetrahedron: $112 \frac{2}{3}$

Height: $\sqrt{26}$

## Problem Kuznetsov Analytic Geometry 7-10