Question
If $\vec{\text{b}}$ is a unit vector such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8,$ find $|\vec{\text{a}}|.$

Answer

Given that $\vec{\text{b}}$ is a unit vector.
$\therefore\big|\vec{\text{b}}\big|=1$
And
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$ (Given)
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow|\vec{\text{a}}|^2-1^2=8$
$\Rightarrow|\vec{\text{a}}|^2=9$
$\therefore|\vec{\text{a}}|=3$

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