91Ó°ÊÓ

• $\stackrel{→}{A}=2.00km,0°ofeast,\stackrel{→}{B}=3.50m,45°southofeast$
  
• The resultant vector of the sailor is $\stackrel{→}{R}=5.80m$,$0°east$

The outcome is a sum total of vector. When you combine two or more vectors, you obtain a new one.

The resultant vector can be given as,

$R=\sqrt{A^2+B^2}$

Here, $A$and $B$are displacement vectors.

The magnitude of third leg of journey can be evaluated as

The magnitude of third leg journey in x-direction,

$C_x=R_x-A_x-B_x$

$C_x=\left(5.80km\right)-\left(2.00km\right)-\left(3.50km\right)\left(\cos 45°\right)$

$C_x=1.33km$

The magnitude of third leg of journey in y-direction

$C_y=R_y-A_y-B_y$

$C_y=\left(0km\right)-\left(0km\right)-\left(-3.50km\right)\left(\sin 45°\right)$

$C_y=2.47km$

The resultant magnitude of third leg of journey is,

$C=\sqrt{{\left(C_x\right)}^2+{\left(C\right)}^{}}$_y2

$C=\sqrt{{\left(1.33km\right)}^2+{\left(2.47km\right)}^2}$

$C=2.81km$

Thus, the magnitude of the third leg is .

The direction of the third leg can be evaluated as,

$\mathrm{θ}={\tan }^{-1}\left(\frac{2.47km}{1.33km}\right)$

$\mathrm{θ}=61.7$north-east

Thus, the third leg lies in the first quadrant.