Vector Algebra Question 117
Question: If $ \overset{\to }{\mathop{a}},,\overset{\to }{\mathop{b}},,\overset{\to }{\mathop{c}}, $ are the position vectors of corners A, B, C of a parallelogram ABCD, then what is the position vector of the corner D?
Options:
A) $ \overset{\to }{\mathop{a}},+\overset{\to }{\mathop{b}},+\overset{\to }{\mathop{c}}, $
B) $ \overset{\to }{\mathop{a}},+\overset{\to }{\mathop{b}},-\overset{\to }{\mathop{c}}, $
C) $ \overset{\to }{\mathop{a}},-\overset{\to }{\mathop{b}},+\overset{\to }{\mathop{c}}, $
D) $ -\overset{\to }{\mathop{a}},+\overset{\to }{\mathop{b}},+\overset{\to }{\mathop{c}}, $
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Answer:
Correct Answer: C
Solution:
- [c] Let O be the origin and ABCD be the parallelogram. In $ \Delta ,ODC, $ $ \overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{CD} $ $ \overrightarrow{CD}=-\overrightarrow{AB} $ and $ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} $ [In $ \Delta ,AOB $ ] $ =\overset{\to }{\mathop{b}},-\overset{\to }{\mathop{a}}, $ Thus, $ \overrightarrow{CD}=-\overrightarrow{AB}=\overrightarrow{a}-\overrightarrow{b} $ So, $ \overrightarrow{OD}=\overrightarrow{c}+\overrightarrow{a}-\overrightarrow{b} $ [since, $ \overrightarrow{OC}=\overrightarrow{C} $ and $ \overrightarrow{CD}=\overrightarrow{a}-\overrightarrow{b} $ ]
BETA
