Question

12. An ant moves 6m towards north
on towards north then it moves 8m to east. How far is ant away from its initial position​

Answer 1

Initially the ant  moves 6m toward north then we can write that as 6i.

Then it moved 8m towards east which we can write as 8i.

So, position vector of the ant = 6i+8j

And the magnitude of  this vector is  √6² + 8²= √100 = 10.

So distance from initial position is 10m.

You can also solve this question by drawing a right angled triangle.

Answer 2

Answer- The above question is from the chapter 'Triangles'.

Let the ant be on O.

Ant moves towards north such that OA= 6 m

Now, this ant moves towards east such that AB= 8 m

Join OB

Clearly, ∠OAB= 90°

We have to find the distance AB i.e. distance of ant from initial point.

By Pythagoras Theorem which states that in a right-angled Δ, sum of squares of two sides is equal to square of the longest side,

AB²= OA² + OB²

AB²= 6² + 8²

AB²= 36 + 64

AB²= 100

AB= ±√100

AB= ± 10 m

AB= 10 m (Rejected -10 m ∵ distance is always positive.)

∴ Distance between ant and initial point= 10 m