Vector Algebra Question 91
Question: Let $ \vec{r}=(\vec{a}\times \vec{b})\sin ,x+(\vec{b}\times \vec{c})\cos ,y+2(\vec{c}\times \vec{a}) $ where $ \vec{a},\vec{b},\vec{c} $ three non-coplanar vectors are. If $ \vec{r} $ is perpendicular to $ \vec{a}+\vec{b}+\vec{c}, $ the minimum value of $ x^{2}+y^{2} $ is
Options:
A) $ {{\pi }^{2}} $
B) $ \frac{{{\pi }^{2}}}{4} $
C) $ \frac{5{{\pi }^{2}}}{4} $
D) None of these
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Answer:
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 $ 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} $
BETA
