Hipi Zhdripi i Matematikës/1158

i cili mund të shkruhet edhe në formë të përcaktorit simbolik të rendit të tretë në këtë mënyrë:

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$ (...25a)

ndërkaq sinusi i këndit ndërmjet atyre vektorëve me formulën:

$sin(\overrightarrow{a},\overrightarrow{b})=\sqrt{\frac{{\left|\begin{array}{cc}{y}_1& z_1\\ y_2& z_2\end{array}\right|}^2+{\left|\begin{array}{cc}{x}_1& z_1\\ x_2& z_2\end{array}\right|}^2+{\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|}^2}{(x_1^2+y_1^2+z_1^2)(x_2^2+y_2^2+z_2^2)}}$ (...26a)

Stampa:Dygishta Ligjet e prodhimit vektorial të vektorëve vërtetohen fare lehtë kur vektorët janë shprehur me koordinata. Për shembull ligji distributiv vërtetohet kështu:

$(\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}$	$=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1+x_2& y_1+y_2& z_1+z_2\\ x_3& y_3& z_3\end{array}\right|$
	$=\overrightarrow{i}\left|\begin{array}{cc}{y}_1+y_2& z_1+z_2\\ y_3& z_3\end{array}\right|-\overrightarrow{j}\left|\begin{array}{cc}{x}_1+x_2& z_1+z_2\\ x_3& z_3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{cc}{x}_1+x_2& y_1+y_2\\ x_3& y_3\end{array}\right|$
	$=(\overrightarrow{i}\left|\begin{array}{cc}{y}_1& z_1\\ y_3& z_3\end{array}\right|-\overrightarrow{j}\left|\begin{array}{cc}{x}_1& z_1\\ x_3& z_3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{cc}{x}_1& y_1\\ x_3& y_3\end{array}\right|+\overrightarrow{k})+(\overrightarrow{i}\left|\begin{array}{cc}{y}_2& z_2\\ y_3& z_3\end{array}\right|-$
	$-j\left|\begin{array}{cc}{x}_2& z_2\\ x_3& z_3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{cc}{x}_2& y_2\\ x_3& y_3\end{array}\right|)$
	$=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_3& y_3& z_3\end{array}\right|+\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}$.

Pra, u vërtetua se $(\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}$.

Stampa:S h e m b u l l i Të njehsohet syprina e paralelogramit të ndërtuar mbi vektorët $\overrightarrow{a}=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ dhe $\overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}$. Stampa:Z g j i d h j e Aplikojmë fomulën (25):

Stampa:Dygishta$S=\overrightarrow{a}+\overrightarrow{b}$	$=\sqrt{{\left|\begin{array}{cc}{y}_1& z_1\\ y_2& z_2\end{array}\right|}^2+{\left|\begin{array}{cc}{x}_1& z_1\\ x_2& z_2\end{array}\right|}^2+{\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|}^2}$
	$=\sqrt{{\left|\begin{array}{cc}1& -1\\ -3& 4\end{array}\right|}^2+{\left|\begin{array}{cc}2& -1\\ 1& 4\end{array}\right|}^2+{\left|\begin{array}{cc}2& 1\\ 1& -3\end{array}\right|}^2}=\sqrt{131}$.