question_answer

If in a triangle \[\overrightarrow{AB}=\mathbf{a},\,\,\overrightarrow{AC}=\mathbf{b}\] and D, E are the mid-points of AB and AC respectively, then \[\overrightarrow{DE}\] is equal to [RPET 1986]

A) \[\frac{\mathbf{a}}{4}-\frac{\mathbf{b}}{4}\]

B) \[\frac{\mathbf{a}}{2}-\frac{\mathbf{b}}{2}\]

C) \[\frac{\mathbf{b}}{4}-\frac{\mathbf{a}}{4}\]

D) \[\frac{\mathbf{b}}{2}-\frac{\mathbf{a}}{2}\]

Correct Answer: D

Solution :

We know by fundamental theorem of proportionality that \[\overrightarrow{DE}=\frac{1}{2}\overrightarrow{BC}\] In triangle, \[\overrightarrow{BC}=\mathbf{b}-\mathbf{a}\]; Hence, \[\overrightarrow{DE}=\frac{1}{2}(\mathbf{b}-\mathbf{a})\].