Answer:
The distance between of the park to the school is approximately 8.6 miles
Step-by-step explanation:
The given information are;
The location of the school from the house = 4 miles East and 1 mile South
The location of the park from the house = 3 miles East and 4 mile South
By representing the direction East (x-direction) as being in the direction of vector [tex]\hat i[/tex] and the direction South (negative y-direction) as being in the direction of vector [tex]\hat j[/tex], we have;
The location of the school from the house = 4·[tex]\hat i[/tex] + [tex]\hat j[/tex]
The location of the park from the house = 3·[tex]\hat i[/tex] + 4·[tex]\hat j[/tex]
The distance, l, between the park and the school is given by the resultant of the two vectors as follows
[tex]l_x[/tex] = 4·[tex]\hat i[/tex] + 3·
[tex]l_y[/tex] = [tex]\hat j[/tex] + 4·
Therefore,
[tex]l = \sqrt{l_x^2 + l_y^2} = \sqrt{7^2 + 5^2} = \sqrt{74} \approx 8.6 \ miles[/tex]
The distance between of the park to the school ≈ 8.6 miles.
