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

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

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

$A_{3}(2 ;-1 ; 5)$

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

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

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

$\overrightarrow{A_{1} A_{3}}=\{2-0 ;-1-(-3) ; 5-1\}=\{2 ; 2 ; 4\}$

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

According to the geometric meaning of the scalar triple product, we have:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)\right|
$$

Let's compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-4 & 4 & 1 \\
2 & 2 & 4 \\
3 & 4 & -5
\end{array}\right|= \\
& =-4 \cdot\left|\begin{array}{cc}
2 & 4 \\
4 & -5
\end{array}\right|-4 \cdot\left|\begin{array}{cc}
2 & 4 \\
3 & -5
\end{array}\right|+1 \cdot\left|\begin{array}{cc}
2 & 2 \\
3 & 4
\end{array}\right|= \\
& =-4 \cdot(-26)-4 \cdot(-22)+1 \cdot 2=104+88+2=194
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|194|=\frac{97}{3}=32 \frac{1}{3}
$$

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:

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

Let's compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-4 & 4 & 1 \\
2 & 2 & 4
\end{array}\right|=i \cdot\left|\begin{array}{cc}
4 & 1 \\
2 & 4
\end{array}\right|-j\left|\begin{array}{cc}
-4 & 1 \\
2 & 4
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-4 & 4 \\
2 & 2
\end{array}\right|= \\
& =i \cdot 14-j \cdot(-18)+k \cdot(-16)=\{14 ; 18 ;-16\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{14^{2}+18^{2}+(-16)^{2}}=\frac{1}{2} \cdot \sqrt{776}=\frac{2 \sqrt{194}}{2}=\sqrt{194}
$$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{97}{3}}{\sqrt{194}}=\frac{97}{\sqrt{97 \cdot 2}}=\sqrt{\frac{97}{2}}$

Volume of the tetrahedron: $32 \frac{1}{3}$

Height: $\sqrt{\frac{97}{2}}$

## Problem Kuznetsov Analytic Geometry 7-26