Vector Algebra Question 56
Question: The vectors $ \vec{a},\vec{b},\vec{c} $ and $ \vec{d} $ are such that $ \vec{a}\times \vec{b}=\vec{c}\times d $ and $ \vec{a}\times \vec{c}=\vec{b}\times \vec{d} $ . Which of the following is/ are correct?
- $ (\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=\vec{0} $
- $ (\vec{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \vec{d})=\vec{0} $
Select the correct answer using the code given below:
Options:
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 nor 2
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ ( \overset{\to }{\mathop{a}},-\overset{\to }{\mathop{d}}, )\times ( \overset{\to }{\mathop{b}},-\overset{\to }{\mathop{c}}, ) $ $ =\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},-\overset{\to }{\mathop{d}},\times \overset{\to }{\mathop{b}},-\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{c}},+\overset{\to }{\mathop{d}},\times \overset{\to }{\mathop{c}}, $ $ =\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}},-\overset{\to }{\mathop{d}},\times \overset{\to }{\mathop{b}},-\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{d}},-\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}}, $ $ =-\overset{\to }{\mathop{d}},\times \overset{\to }{\mathop{b}},+\overset{\to }{\mathop{d}},\times \overset{\to }{\mathop{b}}, $ $ =0 $ Again $ (\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)=(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}},) $ given
$ \Rightarrow (\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)\times (\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}},)=(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}},)\times (\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{d}},)=0 $ $ ( as\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{a}},=0 ) $ So both (1) and (2) are correct.
BETA
