A) \[{{\pi }^{2}}\]
B) \[\frac{{{\pi }^{2}}}{4}\]
C) \[\frac{5{{\pi }^{2}}}{4}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\vec{r}=(\vec{a}\times \vec{b})\sin \,x+(\vec{b}\times \vec{c})\cos \,y+2(\vec{c}\times \vec{a})\] \[\vec{r}.(\vec{a}+\vec{b}+\vec{c})0\] \[\Rightarrow [\vec{a}\vec{b}\vec{c}]\ne 0,\] we have \[\sin x+\cos y=-2\] This is possible only when \[\sin x=-1\] and \[\operatorname{cosy}=-1\]. For \[{{x}^{2}}+{{y}^{2}}\] to be minimum, \[x=\frac{\pi }{2}\] and \[y=\pi \]. \[\therefore \] Minimum value of \[({{x}^{2}}+{{y}^{2}})=\frac{{{\pi }^{2}}}{4}+{{\pi }^{2}}=\frac{5{{\pi }^{2}}}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
