If $\| \mathbf { u } \| = 5$ and $\| \mathbf { v } \| = 3$, and the vectors make angles of $150 ^ { \circ }$ and $110 ^ { \circ }$ with the $x$-axis respectively, find the component form of the sum of $\mathbf { u }$ and $\mathbf { v }$. Round answers to two decimal places.
A) $\langle - 5.36,5.32 \rangle$
B) $\langle - 3.30 , - 0.32 \rangle$
C) $\langle 1.47 , - 1.51 \rangle$
D) $\langle 3.53 , - 7.15 \rangle$
E) $\langle - 4.31,6.20 \rangle$

Question

Quizplus · Accepted Answer

To find the component form of the sum of $\mathbf{u}$ and $\mathbf{v}$, we first find the components of each vector and then add them together. For $\mathbf{u}$, with $\|\mathbf{u}\| = 5$ and making an angle of $150^\circ$ with the x-axis:- $u_x = 5\cos(150^\circ) = 5(-\sqrt{3}/2) = -2.5\sqrt{3}$- $u_y = 5\sin(150^\circ) = 5(1/2) = 2.5$For $\mathbf{v}$, with $\|\mathbf{v}\| = 3$ and making an angle of $110^\circ$ with the x-axis:- $v_x = 3\cos(110^\circ) = 3(-\sqrt{3}/2) \approx -2.598$- $v_y = 3\sin(110^\circ) = 3(\sqrt{3}/2) \approx 2.598$Adding the components of $\mathbf{u}$ and $\mathbf{v}$:- Sum $x$ component: $u_x + v_x = -2.5\sqrt{3} - 2.598 \approx -2.5(1.732) - 2.598 \approx -4.33 - 2.598 \approx -6.928$- Sum $y$ component: $u_y + v_y = 2.5 + 2.598 \approx 5.098$However, the calculation above seems to have an error in interpreting the angles for the sine and cosine functions, leading to incorrect component calculations. Correctly calculating:For $\mathbf{u}$:- $u_x = 5\cos(150^\circ) = 5 \times (-\sqrt{3}/2) \approx -4.33$- $u_y = 5\sin(150^\circ) = 5 \times 1/2 = 2.5$For $\mathbf{v}$:- $v_x = 3\cos(110^\circ) \approx 3 \times (-0.342) \approx -1.026$- $v_y = 3\sin(110^\circ) \approx 3 \times 0.94 \approx 2.82$Summing these components:- Sum $x$ component: $u_x + v_x \approx -4.33 - 1.026 \approx -5.356$- Sum $y$ component: $u_y + v_y \approx 2.5 + 2.82 \approx 5.32$Therefore, the component form of the sum of $\mathbf{u}$ and $\mathbf{v}$ is approximately $\langle -5.36, 5.32 \rangle$, which matches choice A.

If ∥u∥=5\| \mathbf { u } \| = 5∥u∥=5 And ∥v∥=3\| \mathbf { v } \| = 3∥v∥=3

If $\| \mathbf { u } \| = 5$ And $\| \mathbf { v } \| = 3$