Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as
an angle in $$\left[ 0,360 ^ { \circ } \right]$$
-$$\langle 2 \sqrt { 2 } , - 2 \sqrt { 2 } \rangle$$
A) $4 ; 315 ^ { \circ }$
B) $4 \sqrt { 2 } ; 45 ^ { \circ }$
C) $4 ; 135 ^ { \circ }$
D) $8 ; 315 ^ { \circ }$

Question

Quizplus · Accepted Answer

The magnitude of the vector $\langle 2 \sqrt { 2 } , - 2 \sqrt { 2 } \rangle$ is calculated using the formula $\sqrt{x^2 + y^2}$, which gives $\sqrt{(2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$. The direction angle, measured counterclockwise from the positive x-axis, is found by taking the arctan of the y-component divided by the x-component, which gives $\arctan\left(\frac{-2\sqrt{2}}{2\sqrt{2}}\right) = \arctan(-1)$. This corresponds to $225^\circ$, but since we want the angle in $[0,360^\circ]$, we adjust to $315^\circ$.

Find the Magnitude and Direction Angle (To the Nearest Tenth) [0,360∘]\left[ 0,360 ^ { \circ } \right][0,360∘]

Find the Magnitude and Direction Angle (To the Nearest Tenth) $\left[ 0,360 ^ { \circ } \right]$