Question
Find $\left| {\vec a} \right|$ and $\left| {\vec b} \right|$ if $\left( {\vec a + \vec b} \right).\left( {\vec a - \vec b} \right) = 8$ and $\left| {\vec a} \right| = 8\left| {\vec b} \right|$
$\Rightarrow \vec a.\vec a - \vec a.\vec b + \vec b.\vec a - \vec b.\vec b = 8$
$ \Rightarrow {\left| {\vec a} \right|^2} - \vec a.\vec b + \vec a.\vec b - {\left| {\vec b} \right|^2} = 8$
$ \Rightarrow {\left| {\vec a} \right|^2} - \left| {\vec b} \right|^2 = 8$ ...(ii)
Putting ${\left| {\vec a} \right|^2} = 64\left| {\vec b} \right||^2$ in eq. (ii),
$64{\left| {\vec b} \right|^2} - \left| {\vec b} \right|^2 = 8$
$ \Rightarrow \left( {64 - 1} \right){\left| {\vec b} \right|^2} = 8$
$\Rightarrow \left( {63} \right){\left| {\vec b} \right|^2} = 8$
$ \Rightarrow {\left| {\vec b} \right|^2} = \frac{8}{{63}}$
$ \Rightarrow {\left| {\vec b} \right|^2} = \sqrt {\frac{8}{{63}}} = \frac{{2\sqrt 2 }}{{3\sqrt 7 }}$
Putting ${\left| {\vec b} \right|^2} = \frac{{2\sqrt 2 }}{{3\sqrt 7 }}$ in eq (i),
${\left| {\vec a} \right|^2} = 8\left( {\frac{{2\sqrt 2 }}{{3\sqrt 7 }}} \right) = \frac{{16}}{3}\sqrt {\frac{2}{7}} $
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