Direction Tests

Concept:The net displacement from home is the vector sum of all three walks, and the return direction is opposite to that vector.Explanation:Represent each 30 m walk as components using North as $+y$ and East as $+x$.• NW: $(-30/\sqrt{2},\;30/\sqrt{2})$• SW: $(-30/\sqrt{2},\;-30/\sqrt{2})$• SE: $(30/\sqrt{2},\;-30/\sqrt{2})$Sum the x‑components: $(-30-30+30)/\sqrt{2} = -30/\sqrt{2}$Sum the y‑components: $(30-30-30)/\sqrt{2} = -30/\sqrt{2}$So final position is $(-30/\sqrt{2},\;-30/\sqrt{2})$ — southwest of home.To go back home, she must move in the opposite direction: northeast.Answer: North-east