Find the magnitude and positive direction angle of the vector $\langle 3,-3 \sqrt{3}\rangle$ The magnitude of the vector is (Simplify your answer.) The direction angle is (Simplify your answer. Use angle measures greater than or equal to 0 and less than 360 .)

Step 1 ：Given the vector \(\langle 3,-3 \sqrt{3}\rangle\)

Step 2 ：The magnitude of a vector \(\langle a, b \rangle\) is given by \(\sqrt{a^2 + b^2}\)

Step 3 ：Substitute a = 3 and b = -3\(\sqrt{3}\) into the formula, we get the magnitude = 6.0

Step 4 ：The direction angle (in degrees) can be found using the arctangent function, specifically \(\theta = \arctan(\frac{b}{a})\)

Step 5 ：However, since the y-component of the vector is negative, the angle will be in the third or fourth quadrant, so we need to add 180 degrees to the result of the arctangent function to get the correct angle

Step 6 ：Substitute a = 3 and b = -3\(\sqrt{3}\) into the formula, we get the direction angle in radians is -1.0471975511965979

Step 7 ：We need the direction angle in degrees and in the range [0, 360). We can convert the angle to degrees and if the result is negative, add 360 to bring it into the desired range

Step 8 ：Convert the direction angle from radians to degrees, we get the direction angle = 300.0 degrees

Step 9 ：Final Answer: The magnitude of the vector is \(\boxed{6.0}\) and the direction angle is \(\boxed{300.0}\) degrees