Question

plz solve it
plzzzz....

Answer 1

hey mate
here's the solution

Answer 2

$|\vec{A} \times \vec{B}| = \sqrt{3} \: \vec{A} . \vec{B} \\ \\ |\vec{A} || \vec{B}| \sin( \theta) = \sqrt{3} |\vec{A} || \vec{B}| \cos( \theta) \\ \\ \frac{ |\vec{A} || \vec{B}| \sin( \theta)}{ |\vec{A} || \vec{B}| \cos( \theta)} = \sqrt{3} \\ \\ \tan( \theta) = \sqrt{3} \\ \\\tan( \theta) = \tan( {60}^{o} ) \\ \\ \theta = {60}^{o}$

$|\vec{A} + \vec{B}| \\ \\ = \sqrt{ {|\vec{A} | }^{2} + { | \vec{B}|}^{2} +2 |\vec{A} || \vec{B}| \cos( \theta) } \\ \\ = \sqrt{ {|\vec{A} | }^{2} + { | \vec{B}|}^{2} +2 |\vec{A} || \vec{B}| \cos( {60}^{o} ) } \\ \\ = \sqrt{ {|\vec{A} | }^{2} + { | \vec{B}|}^{2} +2 |\vec{A} || \vec{B}| \times \frac{1}{2} } \\ \\ = \sqrt{ {|\vec{A} | }^{2} + { | \vec{B}|}^{2} + |\vec{A} || \vec{B}| } \\ \\ ={({|\vec{A} | }^{2} + { | \vec{B}|}^{2} + |\vec{A} || \vec{B}|)}^{ \frac{1}{2} }$

$\boxed{ \red{ \bold{Answer:\ (4)\ {({|\vec{A} | }^{2} + { | \vec{B}|}^{2} + |\vec{A} || \vec{B}|)}^{ \frac{1}{2} } }}}$