$\text { If }|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=144 ~\&~|\vec{a}|=4 \text { then }|\vec{b}|=$

Using the Lagrange identity for vectors, we have $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2$.

Given that $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144$, we substitute this into the identity:

$|\vec{a}|^2 |\vec{b}|^2 = 144$.

Given $|\vec{a}| = 4$, we have $|\vec{a}|^2 = 16$.

Substituting the value of $|\vec{a}|^2$ into the equation:

$16 |\vec{b}|^2 = 144$.

$|\vec{b}|^2 = \dfrac{144}{16} = 9$.

Taking the square root of both sides, we get $|\vec{b}| = 3$.

Answer: 3