## problem statement

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

$A_{1}(-4 ; 2 ; 6)$

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

$A_{3}(-10 ; 5 ; 8)$

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

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

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

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
6 & -5 & -6 \\
-6 & 3 & 2 \\
-1 & 0 & -10
\end{array}\right|= \\
& =6 \cdot\left|\begin{array}{cc}
3 & 2 \\
0 & -10
\end{array}\right|-(-5) \cdot\left|\begin{array}{cc}
-6 & 2 \\
-1 & -10
\end{array}\right|+(-6) \cdot\left|\begin{array}{cc}
-6 & 3 \\
-1 & 0
\end{array}\right|= \\
& =6 \cdot(-30)+5 \cdot 62-6 \cdot 3=-180+310-18=112
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|112|=\frac{56}{3}=18 \frac{2}{3}
$$

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

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
6 & -5 & -6 \\
-6 & 3 & 2
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-5 & -6 \\
3 & 2
\end{array}\right|-j\left|\begin{array}{cc}
6 & -6 \\
-6 & 2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
6 & -5 \\
-6 & 3
\end{array}\right|= \\
& =i \cdot 8-j \cdot(-24)+k \cdot(-12)=\{8 ; 24 ;-12\}
\end{aligned}
$$

We obtain:

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

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{56}{3}}{14}=\frac{56}{14}=4
$$

Volume of the tetrahedron: $18 \frac{2}{3}$

Height: 4

## Problem Kuznetsov Analytic Geometry $7-2$