Answer:
Let\[\vec{A}=(\overrightarrow{OC})\];\[\vec{B}=(\overrightarrow{OD})\]and
\[\vec{R}=(\overrightarrow{OF})\];
\[\angle
\text{DOF}=\theta \]
Fig. 2(c).54. Resolving \[\vec{B}\] into two rectangular
components, we have: \[\text{B cos}\theta \]along OF and \[\text{B sin}\theta
\]along OE. Here resultant vector is along OF.
\[\text{R}=\text{B cos}\theta \]. As per question,
\[\text{R}=\text{B}/\text{2}=\text{Bcos}\theta
\]
or\[\text{cos}\theta =1/2\]
or\[\theta =\text{6}0{}^\circ \]
Hence, angle between \[\vec{A}\]and \[\vec{B}\]\[\angle
\text{COD }=\text{9}0{}^\circ +\text{6}0{}^\circ =\text{15}0{}^\circ \]
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