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Ta có:→a⋅→b=xaxb+yayb và |→a⋅→b|=|→a||→b|cos(→a;→b)
Lại có:
(→a×→b)2+(→a⋅→b)2
=(xayb−xbya)2+(xaxb+yayb)2
=x2ay2b+x2by2a+x2ax2b+y2ay2b
=(x2a+y2a)(x2b+y2b)
=|→a|2|→b|2
Suy ra: |→a×→b|2=|→a|2|→b|2sin2(→a;→b)
⇒|→a×→b|=|→a||→b|sin(→a;→b), do sin(→a;→b)≥0.
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