the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum? 6…

Answer:

B. $2\sqrt{5}$

Explanation:

Step1: Resolve vectors into components

Let vector $\vec{A}$ with magnitude $A = 4$ at $45^{\circ}$ and vector $\vec{B}$ with magnitude $B = 2$ at $135^{\circ}$. For $\vec{A}$: $A_x=4\cos45^{\circ}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$, $A_y = 4\sin45^{\circ}=2\sqrt{2}$. For $\vec{B}$: $B_x=2\cos135^{\circ}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}$, $B_y = 2\sin135^{\circ}=\sqrt{2}$.

Step2: Find the sum of x - components

$R_x=A_x + B_x=2\sqrt{2}-\sqrt{2}=\sqrt{2}$

Step3: Find the sum of y - components

$R_y=A_y + B_y=2\sqrt{2}+\sqrt{2}=3\sqrt{2}$

Step4: Calculate the magnitude of the resultant vector

$R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(\sqrt{2})^{2}+(3\sqrt{2})^{2}}=\sqrt{2 + 18}=\sqrt{20}=2\sqrt{5}$