Respuesta :
the magnitude and the direction angle for the resultant vector is [tex]3\sqrt{2}miles[/tex] and [tex]\frac{\pi}{4}[/tex] north east .
Step-by-step explanation:
Here we have , Vector u has a magnitude of 3 miles and is directed due east. Vector v has a magnitude of 3 miles and is directed due north. We need to find What are the magnitude and the direction angle for the resultant vector. Let's find out:
We know that for two vectors u , v the resultant vector magnitude is given by :
⇒ [tex]\sqrt{|u|^2+|v|^2+2|u||v|cosx}[/tex] , where x is angle between two vectors
⇒ [tex]\sqrt{(3)^2+(3)^2+(3)(3)cos90}[/tex]
⇒ [tex]\sqrt{9+9+0}[/tex]
⇒ [tex]\sqrt{18}[/tex]
⇒ [tex]3\sqrt{2}miles[/tex]
And , direction is given by :
⇒ [tex]Tan^{-1}(\frac{|v|}{|u|})[/tex]
⇒ [tex]Tan^{-1}(\frac{3}{3})[/tex]
⇒ [tex]Tan^{-1}(1)[/tex]
⇒ [tex]\frac{\pi}{4}[/tex]
Therefore , the magnitude and the direction angle for the resultant vector is [tex]3\sqrt{2}miles[/tex] and [tex]\frac{\pi}{4}[/tex] north east .
