Question

) If |ă + b| = |ā – b| then the angle between å and b
is
​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: If \: |\vec{a} + \vec{b}| = | \vec{a} - \vec{b}| \: then \: angle \: between \: \vec{a} \: and \: \vec{b} \: is$

$\large\underline{\sf{Solution-}}$

We know that

$\: \: \: \: \: \: \: \: \bull \: \: \: \boxed{ \sf{ \: \vec{a}.\vec{a} = { |\vec{a}| }^{2} }}$

$\: \: \: \: \: \: \: \: \bull \: \: \: \boxed{ \sf{ \: \vec{a}.\vec{b} = |\vec{a}| |\vec{b}| cos \theta \: }}$

Now,

Given that,

$\rm :\longmapsto\: |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$

• On squaring both sides, we get

$\rm :\longmapsto\: { |\vec{a} + \vec{b}| }^{2} = { |\vec{a} - \vec{b}| }^{2}$

$\rm :\longmapsto\:(\vec{a} + \vec{b}).(\vec{a} + \vec{b}) = (\vec{a} - \vec{b}).(\vec{a} - \vec{b})$

$\rm :\longmapsto\:\vec{a}.\vec{a} + \vec{a}.\vec{b} + \vec{b}.\vec{a} + \vec{b}.\vec{b} = \vec{a}.\vec{a} - \vec{a}.\vec{b} - \vec{b}.\vec{a} + \vec{b}.\vec{b}$

$\rm :\implies\:2 \: \vec{a}.\vec{b} = - 2 \: \vec{a}.\vec{b}$

$\rm :\longmapsto\:4 \: \vec{a}.\vec{b} = 0$

$\rm :\implies\:\vec{a}.\vec{b} = 0$

$\rm :\longmapsto\: |\vec{a}| |\vec{b}| cos \theta \: = 0$

$\rm :\longmapsto\:cos \theta \: = 0$

$\rm :\implies\: \theta \: = 90 \degree$

Additional Information :-

$\boxed{ \sf{ \: \vec{a}.\vec{b} = \vec{b}.\vec{a}}}$

$\boxed{ \sf{ \: cos \theta \: = \dfrac{\vec{a} \: . \: \vec{b}}{ |\vec{a}| \: |\vec{b}|}}}$

$\boxed{ \sf{ \: \vec{a}. \: \vec{b} = 0 \: \implies\:\vec{a} \: \perp \: \vec{b}}}$

$\boxed{ \sf{ \: \vec{a} \: \perp \: \vec{b} \: then \: \vec{a}. \: \vec{b} \: = 0}}$