To develop this problem we will apply the concepts related to the vector sum. For this purpose, we will take the displacement vectors of the individual and add them to obtain the total distance traveled by him, from the 0 position, to the final position.
[tex]\vec{R} = (+1.0\hat{i}-5\hat{j})+(7\hat{i})+(+12\hat{j})+(6.0\hat{i}+4.0\hat{j})+(+12.0\hat{i}+18.0\hat{j})[/tex]
[tex]\vec{R} = 26\hat{i}+29\hat{j}[/tex]
With the quarterback's position [tex]\vec{Q}[/tex]
[tex]\vec{Q} = -7.0\hat{j}[/tex]
Now we can find the relative position of the two individuals through the subtraction of the vectors, therefore the position between the individual and the quarterback would be
[tex]\vec{R}-\vec{Q} = 26\hat{i}+36\hat{j}[/tex]
Through the properties of the vectors the value of the magnitude of the relative distance would be
[tex]|\vec{R}-\vec{Q}| = \sqrt{26^2+36^2}[/tex]
[tex]|\vec{R}-\vec{Q}| = 44.40yards[/tex]
The direction of the vector would be given by the tangent of the two vectors, that is
[tex]\theta = tan^{-1} (\frac{36}{26}) = 54.16\°[/tex]
Therefore the quarterback will throw the ball 44.4 yards in one direction, with respect to 54.16 °
