mnożenie (x) wektorów; dowod

[kawafis44]

G15.11 Wykazać, że \(\displaystyle{ (a\times b)^2 qslant a^2\cdot b^2}\).

Ja to rozwiązuję tak:

\(\displaystyle{ a=\left[ a_1,a_2,a_3 \right] \\
b=\left[ b_1,b_2,b_3 \right] \\
a b = ft|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right| =
ft[ ft|\begin{array}{cc}a_2&a_3\\b_2&b_3\end{array}\right| , ft|\begin{array}{cc}a_3&a_1\\b_3&b_1\end{array}\right| , ft|\begin{array}{cc}a_1&a_2\\b_1&b_2\end{array}\right| \right] = ft[ a_2 b_3 - b_2 b_3 , a_3 b_1 - b_3 a_1 , a_1 b_2 - b_1 a_2 \right] = \vec{v} \\
(\vec{v})^2 = ? = \vec{v} \circ \vec{v} = (a_2 b_3 - b_2 b_3)^2 + (a_3 b_1 - b_3 a_1)^2 + (a_1 b_2 - b_1 a_2)^2 = ... \\
a^2 \circ b^2 = a_1 b_1 + a_2 b_2 + a_3 b_3 \\
... = (a_2 b_3)^2 -2a_2 b_2 (b_3)^2 + (b_2 b_3)^2 + (a_3 b_1)^2 - 2a_1 a_3 b_1 b_3 + (a_1 b_3)^2 + (a_1 b_2)^2 - 2a_1 a_2 b_1 b_2 + (a_2 b_1)^2 qslant a_1 b_1 + a_2 b_2 + a_3 b_3 \\
a_1 = a , a_2 = b , a_3 = c \\
b_1 = d , b_2 = e , b_3 = f \\
(bf)^2 - 2bef^2 + (ef)^2 + (cd)^2 - 2acdf + (af)^2 + (ae)^2 - 2abde + (bd)^2 qslant ad + be + cf \\
i \ co \ dalej \ ?}\)

pozdrawiam!

[andkom]

Myślę, że jeśli chodzi o \(\displaystyle{ a^2\cdot b^2}\), to chodziło raczej o
\(\displaystyle{ (a\circ a)\cdot(b\circ b)=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}\)