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Question: If in a triangle \(\overset{\rightarrow}{AB} = \mathbf{a},\overset{\rightarrow}{AC} = \mathbf{b}\) a...

If in a triangle AB→=a,AC→=b\overset{\rightarrow}{AB} = \mathbf{a},\overset{\rightarrow}{AC} = \mathbf{b} and D, E are the mid-points of AB and AC respectively, then DE→\overset{\rightarrow}{DE} is equal to

A

a4−b4\frac{\mathbf{a}}{4} - \frac{\mathbf{b}}{4}

B

a2−b2\frac{\mathbf{a}}{2} - \frac{\mathbf{b}}{2}

C

b4−a4\frac{\mathbf{b}}{4} - \frac{\mathbf{a}}{4}

D

b2−a2\frac{\mathbf{b}}{2} - \frac{\mathbf{a}}{2}

Answer

b2−a2\frac{\mathbf{b}}{2} - \frac{\mathbf{a}}{2}

Explanation

Solution

We know by fundamental theorem of proportionality that DE→=12BC→\overset{\rightarrow}{DE} = \frac{1}{2}\overset{\rightarrow}{BC}

In triangle, BC→=b−a\overset{\rightarrow}{BC} = \mathbf{b} - \mathbf{a}; Hence, DE→=12(b−a)\overset{\rightarrow}{DE} = \frac{1}{2}(\mathbf{b} - \mathbf{a}).