In a regular hexagon ABCDEF, AB = a , BC = b and CD = c . Then, AE =
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
In a regular hexagon ABCDEF, \(\overrightarrow{\mathrm{AB}}=\mathbf{a}, \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{c}}\). Then, \(\overrightarrow{\mathrm{AE}}=\)
- \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\)
- \(2 \vec{a}+\vec{b}+\vec{c}\)
- \(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\)
- \(\vec{a}+2 \vec{b}+2 \vec{c}\)
Answer
(C) b + c
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