Question
Let $$\overrightarrow r = \left( {\overrightarrow a \times \overrightarrow b } \right)\sin \,x + \left( {\overrightarrow b \times \overrightarrow c } \right)\cos \,y + 2\left( {\overrightarrow c \times \overrightarrow a } \right)$$ where $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are three non-coplanar vectors. If $$\overrightarrow r $$ is perpendicular to $$\overrightarrow a + \,\overrightarrow b + \,\overrightarrow c $$ then minimum value of $${x^2} + {y^2}$$ is :
A.
$${\pi ^2}$$
B.
$$\frac{{{\pi ^2}}}{4}$$
C.
$$\frac{{5{\pi ^2}}}{4}$$
D.
none of these
Answer :
$$\frac{{5{\pi ^2}}}{4}$$
Solution :
$$\eqalign{
& \overrightarrow r = \left( {\overrightarrow a \times \overrightarrow b } \right)\sin \,x + \left( {\overrightarrow b \times \overrightarrow c } \right)\cos \,y + 2\left( {\overrightarrow c \times \overrightarrow a } \right) \cr
& \overrightarrow r .\left( {\overrightarrow a + \,\overrightarrow b + \,\overrightarrow c } \right) = 0 \cr
& \Rightarrow \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]\left( {\sin \,x + \cos \,y + 2} \right) = 0 \cr} $$
Since, $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] \ne 0,$$ we have $$\sin \,x + \cos \,y = - 2$$
This is possible only when $$\sin \,x = - 1{\text{ and }}\cos \,y = - 1$$
For $${x^2} + {y^2}$$ to be minimum, $$x = \frac{\pi }{2}{\text{ and }}y = \pi $$
$$\therefore $$ Minimum value of $$\left( {{x^2} + {y^2}} \right) = \frac{{{\pi ^2}}}{2} + {\pi ^2} = \frac{{5{\pi ^2}}}{4}$$