Vector Algebra Question 44
Question: If $ \overrightarrow{OA}=\vec{a};\overrightarrow{OB}=\vec{b};\overrightarrow{OC}=2\vec{a}+3\vec{b},; $ $ \overrightarrow{OD}=\vec{a}-2\vec{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{{| \vec{a}\times \vec{b} |}^{2}} $
B) $ 42|\vec{a}\times \vec{b}{{|}^{2}} $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ \overrightarrow{BD}=\overset{\to }{\mathop{a}},-3\overset{\to }{\mathop{b}},,\overrightarrow{AC}=\overset{\to }{\mathop{a}},+3\overset{\to }{\mathop{b}}, $ $ \overrightarrow{BD}\times \overrightarrow{AC}=(\overset{\to }{\mathop{a}},-3\overset{\to }{\mathop{b}},)\times (\overset{\to }{\mathop{a}},+3\overset{\to }{\mathop{b}},)=6\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}}, $ $ \overrightarrow{OD}\times \overrightarrow{OC}=(\overset{\to }{\mathop{a}},-2\overset{\to }{\mathop{b}},)\times (2\overset{\to }{\mathop{a}},+3\overset{\to }{\mathop{b}},)=7\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}}, $ $ (\overrightarrow{BD}\times \overrightarrow{AC}).(\overrightarrow{OD}\times \overrightarrow{OC})=42{{(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)}^{2}} $
BETA
