Vektoriai 3.
$A=\begin{pmatrix}1\text{;} &0\text{;} &5\end{pmatrix}; \\B=\begin{pmatrix}-3\text{;} &0\text{;} &-2\end{pmatrix}; \\C=\begin{pmatrix}-6\text{;} &3\text{;} &-1\end{pmatrix}; \\D=\begin{pmatrix}3\text{;} &-3\text{;} &5\end{pmatrix}; \\$
1)
$\overrightarrow{AD}=\begin{pmatrix}(3-1)\text{;} &(-3-0)\text{;} &(5-5)\end{pmatrix}=\begin{pmatrix}2\text{;} &-3\text{;} &0\end{pmatrix}; \\|\overrightarrow{AD}|=\sqrt{2^2+(-3)^2+0^2}=\sqrt{4+9+0}=\sqrt{13}=3.6056;$
2)
$\overrightarrow{AB}=\begin{pmatrix}(-3-1)\text{;} &(0-0)\text{;} &(-2-5)\end{pmatrix}=\begin{pmatrix}-4\text{;} &0\text{;} &-7\end{pmatrix}; \\|\overrightarrow{AB}|=\sqrt{(-4)^2+0^2+(-7)^2}=\sqrt{16+0+49}=\sqrt{65}=8.0623;$
$\overrightarrow{AC}=\begin{pmatrix}(-6-1)\text{;} &(3-0)\text{;} &(-1-5)\end{pmatrix}=\begin{pmatrix}-7\text{;} &3\text{;} &-6\end{pmatrix}; \\|\overrightarrow{AC}|=\sqrt{(-7)^2+3^2+(-6)^2}=\sqrt{49+9+36}=\sqrt{94}=9.6954;$
Skaliarinė sandauga:
$\overrightarrow{AB}\cdot \overrightarrow{AC}=(-4)\cdot(-7)+0\cdot3+(-7)\cdot(-6)=70;$
Kampas tarp briaunų AB ir AC:
$\varphi=arccos\Biggl(\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}|\cdot|\overrightarrow{AC}|}\Biggr)=arccos\biggl(\frac{70}{8.0623\cdot9.6954}\biggr)=arccos\bigl(0.89552\bigr)=0.46119$
3)
Vektorinė sandauga, pagrindo ABC plotas:
$\overrightarrow{AB}\times \overrightarrow{AC}=|\overrightarrow{AB}|\cdot|\overrightarrow{AC}|\cdot\;sin(\varphi)=8.0623\cdot9.6954\cdot\;sin(0.46119)=\\=sin(\varphi)=8.0623\cdot9.6954\cdot0.44501=34.7851;$
4)
Mišrioji sandauga:
$(\overrightarrow{AB}\times \overrightarrow{AC})\cdot \overrightarrow{AD}=\begin{vmatrix}2 &-3 &0 \\-4 &0 &-7 \\-7 &3 &-6\end{vmatrix}=2\cdot0\cdot(-6)+(-3)\cdot(-7)\cdot(-7)+0\cdot(-4)\cdot3-0\cdot0\cdot(-7)-2\cdot(-7)\cdot3-(-3)\cdot(-4)\cdot(-6)=\\=0+(-147)+0+0+42+72=-33;$
Gretasienio tūris:
$|(\overrightarrow{AB}\times \overrightarrow{AC})\cdot \overrightarrow{AD}|=33$
Plokštumos koeficientai A,B,C:
$A=$
$A_{11}=(-1)^{1+1}\begin{vmatrix}0 &-7 \\3 &-6\end{vmatrix}=(-1)^2\cdot (0\cdot(-6)-3\cdot(-7))=1\cdot21=21;$
$B=$
$A_{12}=(-1)^{1+2}\begin{vmatrix}-4 &-7 \\-7 &-6\end{vmatrix}=(-1)^3\cdot ((-4)\cdot(-6)-(-7)\cdot(-7))=(-1)\cdot(-25)=25;$
$C=$
$A_{13}=(-1)^{1+3}\begin{vmatrix}-4 &0 \\-7 &3\end{vmatrix}=(-1)^4\cdot ((-4)\cdot3-(-7)\cdot0)=1\cdot(-12)=-12;$
$\sqrt{A^2+B^2+C^2}=\sqrt{21^2+25^2+(-12)^2}=\sqrt{441+625+144}=\sqrt{1210}=34.7851;$
Taško atstumas nuo plokštumos, aukštinė:
$h=\cfrac{|(\overrightarrow{AB}\times \overrightarrow{AC})\cdot \overrightarrow{AD}|}{\sqrt{A^2+B^2+C^2}}=\cfrac{33}{34.7851}=0.94868.$