$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 52

If ∣uβƒ—βˆ£=2\bcb{|{ \vec{u}| } = 2} and ∣vβƒ—βˆ£=3\bcb{|{ \vec{v} |} = 3} and uβƒ—\bcb{\vec{u}} has the same direction as vβƒ—\bcb{\vec{v}} then (3uβƒ—)β‹…(2vβƒ—)\bcb{(3\vec{u}) \cdot (2\vec{v})} is:
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