Find the magnitude and the positive direction angle for $\mathbf{u}$. \[ \mathbf{u}=\langle 3,-4\rangle \] \[ |\mathbf{u}|= \] \[ \theta= \] (Round to the nearest tenth as needed. Use angle measures greater than or equal to 0 and less than 360 .)

Step 1 ：Given the vector \(\mathbf{u}=\langle 3,-4\rangle\)

Step 2 ：The magnitude of a vector \(\mathbf{u}=\langle x,y\rangle\) is given by the formula \(|\mathbf{u}|=\sqrt{x^2+y^2}\)

Step 3 ：Substitute \(x = 3\) and \(y = -4\) into the formula, we get \(|\mathbf{u}|=\sqrt{3^2+(-4)^2}=5.0\)

Step 4 ：The direction angle \(\theta\) is given by the formula \(\theta=\arctan(\frac{y}{x})\)

Step 5 ：However, since the y-coordinate is negative, we need to add 180 degrees to the result of the arctan function to get the angle in the correct quadrant

Step 6 ：Substitute \(x = 3\) and \(y = -4\) into the formula, we get \(\theta=\arctan(\frac{-4}{3})+180=126.86989764584402\)

Step 7 ：Round \(\theta\) to the nearest tenth, we get \(\theta=126.9\) degrees

Step 8 ：Final Answer: The magnitude of the vector is \(\boxed{5.0}\) and the positive direction angle is \(\boxed{126.9}\) degrees