Vector Algebra

$\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5$

$\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta $

$\cos \theta = - {3 \over {10}}$

$\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)$

= $6{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}$

= - 118.5

$\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2$      . . . (1)

$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$      . . . (2)

$ \Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$

$ \Rightarrow \,\,\,\tan {\theta \over 2} = n$

$\overrightarrow {{r_g}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$

Let $\overrightarrow C $ = a$\widehat i$ + b$\widehat j$

Given, $\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B $

$ \Rightarrow $ $\,\,\,\,$ a + b = 2 $-$ 1

$ \Rightarrow $ $\,\,\,\,$ a + b = 1 . . . . .(1)

also given

$\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B $

$ \Rightarrow $ $\,\,\,\,$ 2a $-$ b = 1 . . . . (2)

Solving (1) and (2), we get,

a = ${1 \over 3}$ and b = ${2 \over 3}$

$\therefore\,\,\,\,$ $\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$

$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $

= $\sqrt {{5 \over 9}} $

$\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A $

$\overrightarrow A \times \overrightarrow B - \overrightarrow B \times \overrightarrow A = 0$

$ \Rightarrow \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow B = 0$

$\therefore$ $\overrightarrow A \times \overrightarrow B = 0$

$ \Rightarrow AB\sin \theta = 0$

$\theta = $ $0,\pi ,\,\,$$2\pi $ ........

from the given options, $\theta = \pi $