Question

the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum? diagram shows two vectors: one with magnitude 2 at 135° (relative to y-axis or x-axis) and one with magnitude 4 at 45° (relative to x-axis), with coordinate axes. options: 8, 6, 2√5, 20

Explanation:

Step1: Resolve vectors into x and y components

For the vector with magnitude 4 and angle \(45^\circ\) with x - axis:
The x - component \(V_{1x}=4\cos(45^\circ)\), and the y - component \(V_{1y}=4\sin(45^\circ)\). Since \(\cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt{2}}{2}\), we have \(V_{1x}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\) and \(V_{1y}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\).

For the vector with magnitude 2 and angle \(135^\circ\) from the positive x - axis (or \(45^\circ\) from the negative y - axis in terms of reference angle):
The x - component \(V_{2x}=2\cos(135^\circ)\), and \(\cos(135^\circ)=-\frac{\sqrt{2}}{2}\), so \(V_{2x}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\).
The y - component \(V_{2y}=2\sin(135^\circ)\), and \(\sin(135^\circ)=\frac{\sqrt{2}}{2}\), so \(V_{2y}=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\).

Step2: Find the resultant components

The resultant x - component \(R_x = V_{1x}+V_{2x}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\).
The resultant y - component \(R_y = V_{1y}+V_{2y}=2\sqrt{2}+\sqrt{2}=3\sqrt{2}\).

Step3: Calculate the magnitude of the resultant vector

The magnitude of a vector with components \(R_x\) and \(R_y\) is given by \(R = \sqrt{R_x^{2}+R_y^{2}}\).
Substitute \(R_x=\sqrt{2}\) and \(R_y = 3\sqrt{2}\) into the formula:
\[

$$\begin{align*} R&=\sqrt{(\sqrt{2})^{2}+(3\sqrt{2})^{2}}\\ &=\sqrt{2 + 18}\\ &=\sqrt{20}\\ &=2\sqrt{5} \end{align*}$$

\]

Answer:

\(2\sqrt{5}\)