Find |\(\vec{p}\)|,  if for a unit vector \(\vec{a}\),  (\(\vec{p}\)- \(\vec{a}\)).((\(\vec{p}\)+  \(\vec{a}\)) =10

√13 √11  √17 √19

√11

We have    (\(\vec{p}\)- \(\vec{a}\)).((\(\vec{p}\)+  \(\vec{a}\)) =10         ⇒|\(\vec{p}\)|2 - |\(\vec{a}\)|2 =10          ⇒|\(\vec{p}\)|2 - 1  =10    |\(\vec{a}\)|=1 as \(\vec{a}\) is a unit vector)           ⇒|\(\vec{p}\)|2 =11            ⇒|\(\vec{p}\)|= √11