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It is found that\vert \xrightarrow[A]{}+\xrightarrow[B]{}\vert = \vert\xrightarrow[A]{}\vert. This necessarily implies(a) \xrightarrow[B]{} =0(b) \xrightarrow[A]{}, \xrightarrow[B]{} are antiparallel(c) \xrightarrow[A]{}, \xrightarrow[{}]{} are perpendicular(d) \xrightarrow[A]{} .\underset B{→≤0

Answer» It is found that\vert \xrightarrow[A]{}+\xrightarrow[B]{}\vert = \vert\xrightarrow[A]{}\vert. This necessarily implies(a) \xrightarrow[B]{} =0(b) \xrightarrow[A]{}, \xrightarrow[B]{} are antiparallel(c) \xrightarrow[A]{}, \xrightarrow[{}]{} are perpendicular(d) \xrightarrow[A]{} .\underset B{→≤0


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