## 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 $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.

$A_{1}(-3 ; 4 ;-7)$

$A_{2}(1 ; 5 ;-4)$

$A_{3}(-5 ;-2 ; 0)$

$A_{4}(2 ; 5 ; 4)$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{1-(-3) ; 5-4 ;-4-(-7)\}=\{4 ; 1 ; 3\} \\
& \overrightarrow{A_{1} A_{3}}=\{-5-(-3) ;-2-4 ; 0-(-7)\}=\{-2 ;-6 ; 7\} \\
& \overrightarrow{A_{1} A_{4}}=\{2-(-3) ; 5-4 ; 4-(-7)\}=\{5 ; 1 ; 11\}
\end{aligned}
$$

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 will compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
4 & 1 & 3 \\
-2 & -6 & 7 \\
5 & 1 & 11
\end{array}\right|= \\
& =4 \cdot\left|\begin{array}{cc}
-6 & 7 \\
1 & 11
\end{array}\right|-1 \cdot\left|\begin{array}{cc}
-2 & 7 \\
5 & 11
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
-2 & -6 \\
5 & 1
\end{array}\right|= \\
& =4 \cdot(-73)-1 \cdot(-57)+3 \cdot 28=-292-(-57)+84=-151
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-151|=\frac{151}{6}=25 \frac{1}{6}
$$

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 will compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
4 & 1 & 3 \\
-2 & -6 & 7
\end{array}\right|=i \cdot\left|\begin{array}{cc}
1 & 3 \\
-6 & 7
\end{array}\right|-j\left|\begin{array}{cc}
4 & 3 \\
-2 & 7
\end{array}\right|+k \cdot\left|\begin{array}{cc}
4 & 1 \\
-2 & -6
\end{array}\right|= \\
& =i \cdot 25-j \cdot 34+k \cdot(-22)=\{25 ;-34 ;-22\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{25^{2}+(-34)^{2}+(-22)^{2}}=\frac{\sqrt{2265}}{2}
$$

Then

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{151}{6}}{\frac{\sqrt{2265}}{2}}=\frac{151}{\sqrt{15 \cdot 151}}=\sqrt{\frac{151}{15}}
$$

Volume of the tetrahedron: $25 \frac{1}{6}$

Height: $\sqrt{\frac{151}{15}}$

## Problem Kuznetsov Analytic Geometry 7-17