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

\( A_{1}(7 ; 2 ; 4) \)

\( A_{2}(7 ;-1 ;-2) \)

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

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

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

$\overrightarrow{A_{1} A_{2}}=\{7-7 ;-1-2 ;-2-4\}=\{0 ;-3 ;-6\}$

$\overrightarrow{A_{1} A_{3}}=\{3-7 ; 3-2 ; 1-4\}=\{-4 ; 1 ;-3\}$

$\overrightarrow{A_{1} A_{4}}=\{-4-7 ; 2-2 ; 1-4\}=\{-11 ; 0 ;-3\}$

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 compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
0 & -3 & -6 \\
-4 & 1 & -3 \\
-11 & 0 & -3
\end{array}\right|= \\
& =0 \cdot\left|\begin{array}{cc}
1 & -3 \\
0 & -3
\end{array}\right|-(-3) \cdot\left|\begin{array}{cc}
-4 & -3 \\
-11 & -3
\end{array}\right|+(-6) \cdot\left|\begin{array}{cc}
-4 & 1 \\
-11 & 0
\end{array}\right|= \\
& =0 \cdot(-3)+3 \cdot(-21)-6 \cdot 11=0-63-66=-129
\end{aligned}
$$

We obtain:

$V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-129|=\frac{43}{2}=21.5$

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:

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

We compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
0 & -3 & -6 \\
-4 & 1 & -3
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-3 & -6 \\
1 & -3
\end{array}\right|-j\left|\begin{array}{cc}
0 & -6 \\
-4 & -3
\end{array}\right|+k \cdot\left|\begin{array}{cc}
0 & -3 \\
-4 & 1
\end{array}\right|= \\
& =i \cdot 15-j \cdot(-24)+k \cdot(-12)=\{15 ; 24 ;-12\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{15^{2}+24^{2}+(-12)^{2}}=\frac{1}{2} \cdot \sqrt{945}=\frac{3 \sqrt{105}}{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{43}{2}}{\frac{3 \sqrt{105}}{2}}=\frac{43}{\sqrt{105}}$

Volume of the tetrahedron: 21.5

Height: $\frac{43}{\sqrt{105}}$

## Problem Kuznetsov Analytic Geometry 7-3