(a . i)^2+(a . j)^2+(a . k)^2 is equal to:

let  $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$

$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$

so  $(\vec{a} . \hat{i})^2+(\vec{a} . \hat{j})^2+(\vec{o} . \hat{k})^2$

$=a_1^2+a_2^2+a_3^2$

$=|\vec{a}|^2$