Answer to Question #138664 in Mechanics | Relativity for Romelyn Bolo

Question

The resultant of vectors A and B is 12-N perpendicular to vector A. If vector A also has 12-N magnitude, solve for the magnitude and direction of vector B.

Answer 1

Can be seen, that

$|\overrightarrow{B}|=\sqrt{|\overrightarrow {A}|^2+|\overrightarrow{A+B}|^2}$

$|\overrightarrow{B}|=\sqrt{12^2+12^2}=12\sqrt2$ -N

The direction of vector B is 45$\degree$ to vector A and 45$\degree$ to vector A+B cause there is a rectangular triangle.