Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as
an angle in $$\left[ 0,360 ^ { \circ } \right]$$
-$$\langle - 9 \sqrt { 2 } , - 9 \sqrt { 2 } \rangle$$
A) $18 \sqrt { 2 } ; 135 ^ { \circ }$
B) $36 ; 45 ^ { \circ }$
C) $18 ; 225 ^ { \circ }$
D) $18 ; 45 ^ { \circ }$

Question

Quizplus · Accepted Answer

The magnitude of the vector $\langle -9\sqrt{2}, -9\sqrt{2} \rangle$ is calculated using the formula $\sqrt{x^2 + y^2}$, which gives $\sqrt{(-9\sqrt{2})^2 + (-9\sqrt{2})^2} = \sqrt{162 + 162} = \sqrt{324} = 18$. The direction angle, given that both components are negative, places the vector in the third quadrant, where the angle with the positive x-axis is $225^\circ$.

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) $\left[ 0,360 ^ { \circ } \right]$