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

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

$A_{2}(-2 ; 1 ; 0)$

$A_{3}(0 ;-5 ; 1)$

$A_{4}(3 ; 2 ;-6)$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-4 ; 1-(-1) ; 0-3\}=\{-6 ; 2 ;-3\} \\
& A_{1} A_{3}=\{0-4 ;-5-(-1) ; 1-3\}=\{-4 ;-4 ;-2\} \\
& \overrightarrow{A_{1} A_{4}}=\{3-4 ; 2-(-1) ;-6-3\}=\{-1 ; 3 ;-9\}
\end{aligned}
$$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-6 & 2 & -3 \\
-4 & -4 & -2 \\
-1 & 3 & -9
\end{array}\right|= \\
& =-6 \cdot\left|\begin{array}{cc}
-4 & -2 \\
3 & -9
\end{array}\right|-2 \cdot\left|\begin{array}{cc}
-4 & -2 \\
-1 & -9
\end{array}\right|+(-3) \cdot\left|\begin{array}{cc}
-4 & -4 \\
-1 & 3
\end{array}\right|= \\
& =-6 \cdot 42-2 \cdot 34-3 \cdot(-16)=-252-68+48=-272
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-272|=\frac{136}{3}=45 \frac{1}{3}
$$

Since

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

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-6 & 2 & -3 \\
-4 & -4 & -2
\end{array}\right|=i \cdot\left|\begin{array}{cc}
2 & -3 \\
-4 & -2
\end{array}\right|-j\left|\begin{array}{cc}
-6 & -3 \\
-4 & -2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-6 & 2 \\
-4 & -4
\end{array}\right|= \\
& =i \cdot(-16)-j \cdot 0+k \cdot 32=\{-16 ; 0 ; 32\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-16)^{2}+0^{2}+32^{2}}=\frac{1}{2} \cdot \sqrt{1280}=\frac{16 \sqrt{5}}{2}=8 \sqrt{5}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{136}{3}}{8 \sqrt{5}}=\frac{17}{\sqrt{5}}
$$

Volume of the tetrahedron: $45 \frac{1}{3}$

Height: $\frac{17}{\sqrt{5}}$

## Problem Kuznetsov Analytic Geometry 7-19