Afly trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. it will begin and end in the same corner and visit each of the other corners exactly once. to get from a corner to any other corner, it will either fly or crawl in a straight line. what is the maximum possible length, in meters, of its path?

(4 √ 3 + 4 √2 ) m

Step-by-step explanation:

The  insect can travel from one corner directly to opposite corner in four different ways

each can be calculated using Pythagoras theorem

firstly  for one face we need to calculate the diagonal

H² = 1² + 1² = 2

H = √2

then we calculate the diagonal opposite a corner

for example

A to H where A  is at the bottom and H opposite A in another plane at the top in the cubical box

(Interior diagonal for A to H)² = √2² + 1² = 3

Interior diagonal from A to H = √ 3

there are four such corners, the fly will travel 4 √ 3 and it could also go 4 √2 diagonally to the the other corners

maximum possible length in meters = (4 √ 3 + 4 √2 ) m

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this