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
Answers
Answer:
100 Secs
Step-by-step 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
Answer:
100 Secs
Step-by-step 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 Sec