\(\overrightarrow {OA} = 10\overrightarrow k = > A(0;0;10)\)
Ta có: \(OH = OB.\cos 30^\circ = \frac{{15\sqrt 3 }}{2}\)
\(OK = OB.\cos (90^\circ - 30^\circ ) = \frac{{15}}{2}\)
Vậy B(\(\frac{{15}}{2}\);\(\frac{{15\sqrt 3 }}{2}\);0)
=> \(\overrightarrow {AB} = (\frac{{15}}{2};\frac{{15\sqrt 3 }}{2}; - 10)\)
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