1. Start with the two vectors \( \vec{a} \) and \( \vec{b} \) from Part 1 - Graphical Method Example \( \vec{a} \) is a vector of magnitude 15 units at an angle of \( 0^{\circ} \) from the positive, horizontal \( x \)-axis \( \vec{b} \) is a vector of magnitude 10 units at an angle of \( 90^{\circ} \) from the positive, horizontal \( x \)-axis 2. Resolve the vectors into the horizontal x and vertical y components as follows: \[ a_{x}=|a| \cos (\theta) \quad a_{y}=|a| \sin (\theta) \] \( b_{x}=|b| \cos (\theta) \quad b_{y}=|b| \sin (\theta) \) 3. Find the \( \operatorname{components} \) of the resultant vectors as follows: \( s_{x}=a_{x}+b_{x} \) \( s_{y}=a_{y}+b_{y} \) 4. Find the magnitude and angle for the resultant vector as follows: \( |\vec{s}|=\sqrt{\left(s_{x}\right)^{2}+\left(s_{y}\right)^{2}} \) \( \theta=\arctan \left(\frac{s_{y}}{s_{x}}\right) \)

Sure, let's walk through each step to find the resultant vector \( \vec{s} \) by adding vectors \( \vec{a} \) and \( \vec{b} \) using the component method. ### **1. Given Vectors** - **Vector \( \vec{a} \):** - **Magnitude (\( |\vec{a}| \))**: 15 units - **Angle (\( \theta_a \))**: \( 0^{\circ} \) from the positive \( x \)-axis - **Vector \( \vec{b} \):** - **Magnitude (\( |\vec{b}| \))**: 10 units - **Angle (\( \theta_b \))**: \( 90^{\circ} \) from the positive \( x \)-axis ### **2. Resolving Vectors into Components** For each vector, we calculate the horizontal (\( x \)) and vertical (\( y \)) components using trigonometric functions. - **Components of \( \vec{a} \):** \[ \begin{align*} a_x &= |\vec{a}| \cos(\theta_a) = 15 \cos(0^{\circ}) = 15 \times 1 = 15 \\ a_y &= |\vec{a}| \sin(\theta_a) = 15 \sin(0^{\circ}) = 15 \times 0 = 0 \end{align*} \] - **Components of \( \vec{b} \):** \[ \begin{align*} b_x &= |\vec{b}| \cos(\theta_b) = 10 \cos(90^{\circ}) = 10 \times 0 = 0 \\ b_y &= |\vec{b}| \sin(\theta_b) = 10 \sin(90^{\circ}) = 10 \times 1 = 10 \end{align*} \] ### **3. Finding the Components of the Resultant Vector \( \vec{s} \)** Add the corresponding components of \( \vec{a} \) and \( \vec{b} \): \[ \begin{align*} s_x &= a_x + b_x = 15 + 0 = 15 \\ s_y &= a_y + b_y = 0 + 10 = 10 \end{align*} \] So, the components of the resultant vector \( \vec{s} \) are: \[ \vec{s} = \langle 15, 10 \rangle \] ### **4. Calculating the Magnitude and Angle of \( \vec{s} \)** - **Magnitude (\( |\vec{s}| \)):** \[ |\vec{s}| = \sqrt{s_x^2 + s_y^2} = \sqrt{15^2 + 10^2} = \sqrt{225 + 100} = \sqrt{325} \approx 18.03 \text{ units} \] - **Angle (\( \theta_s \)):** \[ \theta_s = \arctan\left(\frac{s_y}{s_x}\right) = \arctan\left(\frac{10}{15}\right) = \arctan\left(\frac{2}{3}\right) \approx 33.69^{\circ} \] ### **5. Summary of the Resultant Vector** - **Components:** \( \langle 15, 10 \rangle \) - **Magnitude:** Approximately **18.03 units** - **Direction:** Approximately **\( 33.69^{\circ} \)** above the positive \( x \)-axis ### **Visual Representation** To better understand the resultant vector, here's a graphical representation: ```plaintext y | | * (10 units) | / | / | / | / | / | / | / θ_s ≈ 33.69° | / |/__________ x 15 units ``` - **Vector \( \vec{a} \):** Along the positive \( x \)-axis (15 units). - **Vector \( \vec{b} \):** Along the positive \( y \)-axis (10 units). - **Resultant Vector \( \vec{s} \):** Diagonal from the origin to the point \( \langle 15, 10 \rangle \). This approach effectively breaks down the vectors into their components, sums them, and then recombines them to find the magnitude and direction of the resultant vector.