Answer
The quaterback should throw the ball to a distance of $38.5yd$ in the direction $24.6^{\circ}$ to the right of downfield from his location.
Work Step by Step
We take the position of the ball before the play starts as the origin of our coordinate system.
Let $A$ be the postion vector of the receiver at start and $B,C,D,E$ be the subsequent displacement vector of the receiver.
Also, let $P$ be the resultant positon vector of the receiver after the displacements.
$A=1.0i-5.0j$, $B=9.0i$, $C=11.0j$, $D=-6.0i+4.0j$, $E=12.0i+18.0j$
We know, $P=A+B+C+D+E=(1.0+9.0-6.0+12)i+(-5.0+11.0+4.0+18)j=16.0i+28.0j$
Again, let $Q$ be the position vector of the quaterback and $R$ be the displacement vector from the quaterback to the receiver.
Now, $Q=-7.0j$
and $R=P-Q=16.0i+35.0j$
Then, the magnitude and angle is given by
$|R|=\sqrt (16.0^2+35.0^2)=38.5yd$
and $\theta=tan^{-1}(\frac{35.0}{16.0})=65.4^{\circ}$ to the downfield of right or $24.6^{\circ}$ to the right of downfield
Thus, the quaterback should throw the ball to a distance of $38.5yd$ in the direction $24.6^{\circ}$ to the right of downfield from his location.
