Question

the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum?
options: 20, 6, $2\sqrt{5}$, 8

Explanation:

Step1: Resolve vectors into x and y components

For the vector with magnitude 2 and angle \(135^\circ\) (from positive x - axis):
The x - component \(v_{1x}=2\cos135^\circ\), and \(\cos135^\circ =-\frac{\sqrt{2}}{2}\), so \(v_{1x}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\).
The y - component \(v_{1y}=2\sin135^\circ\), and \(\sin135^\circ=\frac{\sqrt{2}}{2}\), so \(v_{1y}=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\).

For the vector with magnitude 4 and angle \(45^\circ\) (from positive x - axis):
The x - component \(v_{2x}=4\cos45^\circ\), and \(\cos45^\circ=\frac{\sqrt{2}}{2}\), so \(v_{2x}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\).
The y - component \(v_{2y}=4\sin45^\circ\), and \(\sin45^\circ=\frac{\sqrt{2}}{2}\), so \(v_{2y}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\).

Step2: Find the resultant x and y components

The resultant x - component \(V_x = v_{1x}+v_{2x}=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\).
The resultant y - component \(V_y = v_{1y}+v_{2y}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\).

Step3: Calculate the magnitude of the resultant vector

The magnitude of a vector \(\vec{V}\) with components \(V_x\) and \(V_y\) is given by \(|\vec{V}|=\sqrt{V_x^{2}+V_y^{2}}\).
Substitute \(V_x = \sqrt{2}\) and \(V_y = 3\sqrt{2}\) into the formula:
\[

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

\]

Answer:

\(2\sqrt{5}\)