Question

Rite a formula for a two-dimensional vector field which has all vectors of length 8 and perpendicular to the position vector at that point.

Answer 1

Hello,

Maybe i am wrong.

Let's assume (x,y) the position vector u
$\overrightarrow{u}=x*\overrightarrow{i}+y*\overrightarrow{j}\\\\\overrightarrow{v}=v_1*\overrightarrow{i}+v_2*\overrightarrow{j}$
$\sqrt{v_1^2+v_2^2} =8 ==\ \textgreater \ v_1^2+v_2^2=64\\\\ 0=\overrightarrow{u}.\overrightarrow{v}=x*v_1+y*v_2\\\\ v_2= \dfrac{-x}{y} *v_1\\\\ v_1^2+ \dfrac{x^2}{y^2} =64\\\\ v_1^2* \dfrac{x^2+y^2}{y^2} =64\\\\ v_1=\pm8* \dfrac{y}{ \sqrt{x^2+y^2} } v_2=-\pm8* \dfrac{x}{ \sqrt{x^2+y^2} }$