here are vectors $vec{v}$ and $vec{w}$. find the magnitude and direction angle ($\theta$) of…

Explanation:

Step1: Convert vectors to component - form

For vector $\vec{v}$ with magnitude $|\vec{v}| = 3$ and direction $\theta_v=310^{\circ}$: $v_x = |\vec{v}|\cos\theta_v=3\cos310^{\circ}\approx3\times0.6428 = 1.93$ $v_y = |\vec{v}|\sin\theta_v=3\sin310^{\circ}\approx3\times(- 0.7660)=-2.30$

For vector $\vec{w}$ with magnitude $|\vec{w}| = 8$ and direction $\theta_w = 190^{\circ}$: $w_x=|\vec{w}|\cos\theta_w=8\cos190^{\circ}\approx8\times(-0.9848)= - 7.88$ $w_y=|\vec{w}|\sin\theta_w=8\sin190^{\circ}\approx8\times(-0.1736)=-1.39$

Step2: Find the components of $\vec{v}+\vec{w}$

$(\vec{v}+\vec{w})_x=v_x + w_x=1.93-7.88=-5.95$ $(\vec{v}+\vec{w})_y=v_y + w_y=-2.30 - 1.39=-3.69$

Step3: Calculate the magnitude of $\vec{v}+\vec{w}$

$|\vec{v}+\vec{w}|=\sqrt{(\vec{v}+\vec{w})_x^2+(\vec{v}+\vec{w})_y^2}=\sqrt{(-5.95)^2+(-3.69)^2}=\sqrt{35.4025 + 13.6161}=\sqrt{49.0186}\approx7.0$

Step4: Calculate the direction angle of $\vec{v}+\vec{w}$

$\tan\theta=\frac{(\vec{v}+\vec{w})_y}{(\vec{v}+\vec{w})_x}=\frac{-3.69}{-5.95}\approx0.62$ Since both $x$ - component and $y$ - component are negative, the vector is in the third quadrant. $\theta=\arctan(0.62)+180^{\circ}\approx31.8^{\circ}+180^{\circ}=211.8^{\circ}$

Answer:

Magnitude: $7.0$, Direction angle: $211.8^{\circ}$