An aeroplane flies 400 m due north and then 300 m due south and then flies 1200 m upwards, the net displacement is ?

The net displacement of the plane will be the shortest distance between its starting point and its finish point.

Let's assume that your plane starts in point `A`. The first stage of its flight will take it 400 m north to point `B`. The second stage of its flight will take it 300 m south to point `C`.

Since north and south are opposite directions, the plane will end up very close to where it started, i.e. point `A`. More specifically, point `AC` will be

`underbrace(AC)_(color(blue)("horizontal displacement")) = AB - BC = 400 - 300 = "100 m"`

So, horizontally, the plane only travelled 100 m from its starting point. At this point, its flight path turns upwards.

The plane flies 1200 m upwards to point `D`, which is its finish point. The distance between point `A` and point `D` can be seen as the hypothenuse of a right triangle with sides 100 and 1200.

This means that the net displacement, which is equal to `AD`, will be

`(AD)^2 = underbrace((AC)^2)_(color(blue)("horiz displacement")) + overbrace((CD)^2)^(color(green)("vert displacement"))`

`(AD)^2 = 100""^2 + 1200""^2`

`AD = sqrt(100""^2 + 1200""^2) ~= color(green)("1204 m")`