A) \[-\frac{1}{2}\]
B) \[-\frac{3}{2}\]
C) \[-1\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[\vec{a}=\hat{i}-2\hat{j}+\hat{k}\,;\,\,\vec{b}=\hat{i}-\hat{j}+\hat{k}\] \[\because \,\,\,\,\,\,\vec{b}\times \vec{c}=\vec{b}\times \vec{a}\,\,\,\,\,\Rightarrow \text{ }\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\vec{a}\times \left( \vec{b}\times \vec{a} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\left( \vec{a}.\vec{c} \right)\vec{b}-\left( \vec{a}.\vec{b} \right)\vec{c}=\left( \vec{a}.\vec{a} \right)\vec{b}-\left( \vec{a}.\vec{b} \right)\vec{a}\] \[\because \,\,\,\,\,\,~\vec{a}.\vec{c}=0\] \[\Rightarrow \,\,\,\,\,\,\vec{c}=-\frac{1}{2}\left( \hat{i}+\hat{j}+\hat{k} \right)\] \[\therefore \,\,\,\,\,\,\vec{b}.\vec{c}=\frac{-1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
