A bird is at pont P whose coordinates are(4m,-1m,5m).The bird observes two points P1 P2 having coordinates(-1m,2m,0m,)and (1m,1m,4m)respectively.At time t=0,it starts flying in the plane of three positions with the constant speed of 5m/sec in a direction perpendicular to the straight line P1,P2 till it sees P1 and P2 collinear at time t.Calulate t.
Equation of plane with three points
x-x1
y-y1
z-z1
= 0
x2-x1
y2-y1
z2-z1
x3-x1
y3-y1
z3-z1
x- 4
y- (-1)
z -5
= 0
(-1) - 4
2 - (-1)
0 - 5
1 - 4
1 - (-1)
4 - 5
x -4
y -(-1)
z -5
= 0
-5
3
-5
-3
2
-1
(x -
4
)(
3
·
(-1)
-
(-5)
·
2
) - (y -
(-1)
)(
(-5)
·
(-1)
-
(-5)
·
(-3)
) + (z -
5
)(
(-5)
·
2
-
3
·
(-3)
) = 0
7
(x-
4
)+
10
(y-
(-1)
)+
(-1)
(z-
5
) = 0
2x-y+4z=29
also it satisfies the plane so
7x+10y-z=13
solvin all three eq.
x=15/7,y=3/7,z=44/7
distance =
divide by velocity to get the answer
7 x + 10 y - z - 13 =0 a - b ={ ax - bx ; ay - by ; az - bz }={(-1)-1;2-1;0-4}={-2;1;-4} Lets a x,y,z lies on the line perpendicular to AB and in the plane then x-4,y+1,z-5 will also lie in the plane and perpendicular to AB where x,y,z is location of bird at any time. dot product of these two point should be zero =>8x+4y-3z=0 since at some time it becomes collinear with AB => triangle formed be A,B and that point will be zero =>
