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

$A_{1}(1 ; 3 ; 0)$

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

$A_{3}(3 ; 0 ; 1)$

$A_{4}(-4 ; 3 ; 5)$
## Solution

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

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

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

$\overrightarrow{A_{1} A_{4}}=\{-4-1 ; 3-3 ; 5-0\}=\{-5 ; 0 ; 5\}$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
3 & -4 & 2 \\
2 & -3 & 1 \\
-5 & 0 & 5
\end{array}\right|= \\
& =3 \cdot\left|\begin{array}{cc}
-3 & 1 \\
0 & 5
\end{array}\right|-(-4) \cdot\left|\begin{array}{cc}
2 & 1 \\
-5 & 5
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
2 & -3 \\
-5 & 0
\end{array}\right|= \\
& =3 \cdot(-15)+4 \cdot 15+2 \cdot(-15)=-45+60-30=-15
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-15|=\frac{5}{2}=2.5
$$

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:

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

We compute the vector product:

$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
3 & -4 & 2 \\
2 & -3 & 1
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-4 & 2 \\
-3 & 1
\end{array}\right|-j\left|\begin{array}{cc}
3 & 2 \\
2 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
3 & -4 \\
2 & -3
\end{array}\right|= \\
& =i \cdot 2-j \cdot(-1)+k \cdot(-1)=\{2 ; 1 ;-1\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{2^{2}+1^{2}+(-1)^{2}}=\frac{\sqrt{6}}{2}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{5}{2}}{\frac{\sqrt{6}}{2}}=5 \cdot \frac{3}{\sqrt{6}}=5 \sqrt{\frac{3}{2}}
$$

Volume of the tetrahedron: 2.5

Height: $\sqrt[5]{\frac{3}{2}}$

## Problem Kuznetsov Analytic Geometry 7-27