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

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

$A_{2}(-1 ; 1 ; 3)$

$A_{3}(2 ;-2 ; 4)$

$A_{4}(-1 ; 0 ;-2)$
## Solution

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

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

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

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

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-2 & 0 & 1 \\
1 & -3 & 2 \\
-2 & -1 & -4
\end{array}\right|= \\
& =-2 \cdot\left|\begin{array}{cc}
-3 & 2 \\
-1 & -4
\end{array}\right|-0 \cdot\left|\begin{array}{cc}
1 & 2 \\
-2 & -4
\end{array}\right|+1 \cdot\left|\begin{array}{cc}
1 & -3 \\
-2 & -1
\end{array}\right|= \\
& =-2 \cdot(14)-0+1 \cdot(-7)=-28-7=-35
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-35|=\frac{35}{6}=5 \frac{5}{6}
$$

Since

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

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:

$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-2 & 0 & 1 \\
1 & -3 & 2
\end{array}\right|=i \cdot\left|\begin{array}{cc}
0 & 1 \\
-3 & 2
\end{array}\right|-j\left|\begin{array}{cc}
-2 & 1 \\
1 & 2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-2 & 0 \\
1 & -3
\end{array}\right|= \\
& =3 \cdot i+5 \cdot j+6 \cdot k=\{3 ; 5 ; 6\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{3^{2}+5^{2}+6^{2}}=\frac{1}{2} \cdot \sqrt{70}=\frac{\sqrt{70}}{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{35}{6}}{\frac{\sqrt{70}}{2}}=\frac{35}{\sqrt{70}}=\frac{35}{\sqrt{2} \cdot \sqrt{35}}=\sqrt{\frac{35}{2}}$

Volume of the tetrahedron: $5 \frac{5}{6}$

Height: $\sqrt{\frac{35}{2}}$

## Problem Kuznetsov Analytic Geometry 7-13