Find the magnitude and direction of \(-\vec{A}+\vec{B},\) where \(\vec{A}=(23.0,59.0), \vec{B}=(90.0,-150.0)\)

First, we need to find the components of the vector \(-\vec{A}+\vec{b}\): $$ -\vec{A}+\vec{B} = (-\vec{A}_x,\,-\vec{A}_y) + (\vec{B}_x,\,\vec{B}_y) $$ where \(\vec{A}=(\vec{A}_x, \vec{A}_y)\) and \(\vec{B}=(\vec{B}_x,\,\vec{B}_y)\) Given: $$ \vec{A} = (23.0, 59.0)\quad and \quad \vec{B} = (90.0, -150.0) $$ then the resultant vector is: $$ -\vec{A}+\vec{B} = (-23.0 - 90.0, -59.0 + 150.0) = (-113.0, 91.0) $$

Now that we have the components of the resulting vector, we can find its magnitude using the Pythagorean theorem: $$ |\vec{R}| = \sqrt{(-113.0)^2 + (91.0)^2} \approx 145.18 $$

Finally, we need to find the direction of the resultant vector. We can use the inverse tangent function (arctangent) to find the angle with respect to the positive x-axis. The formula for the direction \(\theta\) from the components of the vector is given by: $$ \theta = \arctan \left(\frac{R_y}{R_x}\right) $$ For our vector, this will be: $$ \theta = \arctan \left(\frac{91.0}{-113.0}\right) $$ Computing the arctangent: $$ \theta \approx -39.08^\circ $$ If we want to find the angle in the standard polar notation (positive angle measured counterclockwise from the positive x-axis), we can add \(360^\circ\) to our result: $$ \theta = -39.08^\circ + 360^\circ = 320.92^\circ $$ Our final answer is that this vector has a magnitude of approximately 145.18 units and a direction of approximately 320.92 degrees.