Vector Algebra Question 67
Question: If $ \overset{\to }{\mathop{a}},,,\overset{\to }{\mathop{b}},,,\overset{\to }{\mathop{c}}, $ are three non-coplanar vectors, then the value of $ \frac{\overset{\to }{\mathop{a}},.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},)}{(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{a}},).\overset{\to }{\mathop{b}},}+\frac{\overset{\to }{\mathop{b}},.(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{c}},)}{\overset{\to }{\mathop{c}},.(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)} $ is:
Options:
A) 0
B) 2
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] By definition of scalar triple product $ \vec{a}\cdot (\vec{b}\times \vec{c}) $ can be written as $ [\vec{a}\vec{b}\vec{c}] $ $ \frac{\vec{a}.(\vec{b}\times \vec{c})}{(\vec{c}\times \vec{a}).\vec{b}}+\frac{\vec{b}.(\vec{a}\times \vec{c})}{\vec{c}.(\vec{a}\times \vec{b})}=\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{c}\vec{a}\vec{b}]}+\frac{[\vec{b}\vec{a}\vec{c}]}{[\vec{c}\vec{a}\vec{b}]} $ $ =\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{a}\vec{b}\vec{c}]}-\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{a}\vec{b}\vec{c}]}=1-1=0 $ $ \because [\vec{a}\vec{b}\vec{c}]=[\vec{b}\vec{c}\vec{a}]=[\vec{c}\vec{a}\vec{b}] $ but $ [\vec{b}\vec{c}\vec{a}]=-[ \vec{a}\vec{b}\vec{c} ] $
BETA
