Question

A bird at a point (4,-1,-5) sees two point (-1,-1,0 and p2(3,-1,-3).At time t =0 it starts flying with a constant speed of 10 m/s to be in line pi and p2 in miminmum possible time. Find t if all coordinates are in km.

Answer 1

Answer:

100 sec

Explanation:

A bird is at a point p(4,-1,-5) and sees two points p1 (-1,-1,0) and p2(3,-1,-3) . at time t=0 it starts flying with a constant speed of 10m/s to be in line with points p1 and p2 in minimum possible time t . finde t , if all coordinates are in km

y coordinate is fixed -1 so we will ignore that

to be in line with p1 (-1,-1,0) and p2(3,-1,-3)

Slope of line = (z₂ - z₁)/(x₂ - x₁)

= (-3 - 0)/(3 -(-1))

= -3/4

z = -3x/4 + c

=> 0 = 3/4 + c

=> c = -3/4

z = -3x/4 -3/4

=> 4z = -3x -3

=>  4z + 3x = -3   - eq 1

To come in line at earliest Bird need to come perpendicularly

slope of perpendicular line = m

m(-3/4) = -1

=> m = 4/3

z = 4x/3 + c

=> -5 = 4*4/3 + c

=> -15 = 16 + 3c

=> c = -31/3

=> z = 4x/3 - 31/3

=> 3z - 4x = - 31   - eq 2

4 * eq 1 + 3 * eq 2

25z = -105

z = -105/25

z = -21/5

4 * (-21/5)  + 3x = - 3

=> -84  + 15x = -15

=> 15x = 69

=> x = 69/15

=> x = 23/5

(23/25   ,  -1  , -21/5) will be the point in line of p1 & p2

Distance from p(4,-1,-5)

= √(4 - 23/5)² + 0² + (-5 -(-21/5))²

= √(-3/5)² + (4/5)²

= √9/25 + 16/25

= √25/25

= 1

Distance covered = 1 km = 1000 m

Speed = 10 m/s

Time = 1000/10 = 100 Secs