A) 0
B) 2
C) 1
D) None of these
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 \[[\vec{b}\vec{c}\vec{a}]=-\left[ \vec{a}\vec{b}\vec{c} \right]\]You need to login to perform this action.
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