Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as
an angle in [0,360°].
-$\langle 12 \sqrt { 2 } , - 12 \sqrt { 2 } \rangle$
A) $24 ; 315 ^ { \circ }$
B) $24 ; 135 ^ { \circ }$
C) $24 \sqrt { 2 } ; 45 ^ { \circ }$
D) $48 ; 315 ^ { \circ }$

Question

Quizplus · Accepted Answer

The magnitude of the vector $\langle 12 \sqrt { 2 } , - 12 \sqrt { 2 } \rangle$ is calculated using the formula $\sqrt{x^2 + y^2}$, which gives $\sqrt{(12\sqrt{2})^2 + (-12\sqrt{2})^2} = \sqrt{288 + 288} = \sqrt{576} = 24$. The direction angle, given that the vector points in the fourth quadrant (positive x, negative y), is $360^\circ - 45^\circ = 315^\circ$.

Find the Magnitude and Direction Angle (To the Nearest Tenth) ⟨122,−122⟩\langle 12 \sqrt { 2 } , - 12 \sqrt { 2 } \rangle⟨122​,−122​⟩

Find the Magnitude and Direction Angle (To the Nearest Tenth) $\langle 12 \sqrt { 2 } , - 12 \sqrt { 2 } \rangle$