Vector cross product
definition
\ [\ Bf and \ times \ vec b = | \ thing and || \ bf b | sin \ theta \]
prove
- Proof: In the parallel quadrilateral 0ACB as shown \ [S _ {\ Delta AOC } = \ frac {1} {2} | \ vec {a} || \ vec b | sin \ theta \]
- The area of a parallelogram is \ [S = | \ VEC {A} | | \ VEC B | SiN \ Theta \]
\ [\ VEC A \ CDOT \ VEC B = | \ VEC A | | \ VEC B | COS \ Theta \]
\ [COS \ Theta = \ FRAC {\ VEC A \ CDOT \ VEC {B} | \ A VEC | | \ B VEC |} \]
\[ \begin{eqnarray} sin\theta &= & \sqrt{1-cos^2\theta} \\ &=&\frac{\sqrt{(|\vec a|^2\cdot|\vec b|)^2-(\vec{a}\cdot{\vec{b})^2}}}{|\vec a||\vec b|} \\ \end{eqnarray} \]
\begin{eqnarray}
S &=& \sqrt{(|\vec a|^2\cdot|\vec b|)^2-(\vec{a}\cdot{\vec{b})^2}} \
&=& \sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)-(x_1x_2+y_1y_2)^2} \
&=& \sqrt{(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2} \
&=& \sqrt{(x_1y_2-x_2y_1)^2} \
&=& |x_1y_2-x_2y_1| \
\end{eqnarray}