Question

Find $|\overrightarrow{b}|$, if $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8$ and $|\overrightarrow{a}|=8|\overrightarrow{b}|$

Answer 1

$(\overrightarrow{a}\cdot \overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=8$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{a}-\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}-\overrightarrow{b}\cdot \overrightarrow{b}=8$$

$\Rightarrow |\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=8$

$\Rightarrow (8|\overrightarrow{b}|{)}^2-|\overrightarrow{b}{|}^2=8,[\because |\overrightarrow{a}|=8|\overrightarrow{b}|]$

$\Rightarrow 64|\overrightarrow{b}{|}^2-|\overrightarrow{b}{|}^2=8$

$\Rightarrow 63|\overrightarrow{b}{|}^2=8$

$\Rightarrow |\overrightarrow{b}{|}^2=\frac{8}{63}$

$\Rightarrow |\overrightarrow{b}|=\sqrt{\frac{8}{63}}$ [Magnitude of a vector is non-negative]

$\therefore |\overrightarrow{b}|=\frac{2\sqrt{2}}{3\sqrt{7}}$ and $|\overrightarrow{a}|=\frac{16\sqrt{2}}{3\sqrt{7}}$