Question

review the graph
what are the characteristics of the vector shown?
magnitude of 8 and direction angle equal to 0°
magnitude of 8 and direction angle equal to 180°
magnitude of 2√2 and direction angle equal to 0°
magnitude of 2√2 and direction angle equal to 180°

Explanation:

Step1: Calculate the magnitude

The vector \(u\) starts at \((- 2,0)\) and ends at \((6,0)\). The formula for the magnitude of a vector \(\vec{v}=(x_2 - x_1,y_2 - y_1)\) is \(|\vec{v}|=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(x_1=-2,y_1 = 0,x_2 = 6,y_2=0\). Then \(|\vec{u}|=\sqrt{(6-(-2))^2+(0 - 0)^2}=\sqrt{8^2}=8\).

Step2: Determine the direction angle

Since the vector lies on the positive \(x -\)axis. The direction angle \(\theta\) of a vector \(\vec{v}=(x,y)\) is given by \(\tan\theta=\frac{y}{x}\) (when \(x
eq0\)). Here \(y = 0\) and \(x>0\), so \(\theta = 0^{\circ}\)

Answer:

magnitude of \(8\) and direction angle equal to \(0^{\circ}\)