Vector Algebra Question 103
Question: If $ \overrightarrow{OA}=\overrightarrow{a};,\overrightarrow{OB}=\overrightarrow{b};,\overrightarrow{OC}=2\overrightarrow{a}+3\overrightarrow{b}; $ $ \overrightarrow{OD}=\overset{\to }{\mathop{a}},-2\overset{\to }{\mathop{b}},, $ the length of $ \overrightarrow{OA} $ is three times the length of $ \overrightarrow{OB} $ and $ \overrightarrow{OA} $ is perpendicular to $ \overrightarrow{DB} $ then $ (\overrightarrow{BD}\times \overrightarrow{AC}).(\overrightarrow{OD}\times \overrightarrow{OC}) $ is
Options:
A) $ 7|\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},{{|}^{2}} $
B) $ 42|\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},{{|}^{2}} $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ \overrightarrow{BD}=\overrightarrow{a}-3\overrightarrow{b},\overrightarrow{AC}=\overrightarrow{a}+3\overrightarrow{b} $ $ \overrightarrow{BD}\times \overrightarrow{AC}=(\overrightarrow{a}-3\overrightarrow{b})\times (\overrightarrow{a}+3\overrightarrow{b})=6\overrightarrow{a}\times \overrightarrow{b} $ $ \overrightarrow{OD}\times \overrightarrow{OC}=(\overrightarrow{a}-2\overrightarrow{b})\times (2\overrightarrow{a}+3\overrightarrow{b})=7\overrightarrow{a}\times \overrightarrow{b} $ $ (\overrightarrow{BD}\times \overrightarrow{AC}).(\overrightarrow{OD}\times \overrightarrow{OC})=42{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}} $
BETA
