Answer:
The terminal point is Q(0,4)
Step-by-step explanation:
We are given the following in the question:
Initial point of vector, P(1,0)
The terminal point Q lies on the y-axis, thus Q have coordinates of the form:
[tex]Q(0,y)[/tex]
The magnitude of the vector is [tex]\sqrt{65}[/tex]
Magnitude is given by:
[tex]v = \sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\\\sqrt{65}=\sqrt{(0-1)^2+(y-0)^2}\\\sqrt{65}=\sqrt{1+y^2}\\65 = 1+y^2\\y^2 = 64\\y = \pm 8[/tex]
But since the terminal point is above the initial point, thus,
[tex]y > 0\\\Rightarrow y = 4[/tex]
Thus, the terminal point is Q(0,4)
