Vector Algebra Question 255
Question: 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]
Options:
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} $
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Answer:
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}) $ .
BETA
