Here positions of three points in 3D are given as (taking unit of length as #m#)
#P->(4,-1,5)#
#P_1->(-1,-1,0)#
#P_2->(3,-1,-3)#
So
#vec(PP_1)=-5hati-5hatk#
#vec(PP_2)=-hati-8hatk#
#vec(P_1P_2)=4hati-3hatk#
If the positon of bird ,when it is on line #P_1P_2# be #O#.
1st method
Let #OP_1:OP_2=m:n#
So #vec(PO)=(n*vec(PP_1)+m*vec(PP_2))/(m+n)#
#=>vec(PO)=(n(-5hati-5hatk)+m(-hati-8hatk))/(m+n)#
#=>vec(PO)=((-5n-m)hati+(-5n-8m)hatk)/(m+n)#
Now #vec(PO) and vec(P_1P_2)# are perpendicularly inclined.
So #vec(PO) *vec(P_1P_2)=0#
#=>(4(-5n-m)-3(-5n-8m))/(m+n)=0#
#=>-20n-4m+15n+24m=0#
#=>20m=5n#
#=>m/n=1/4#
~~~~~~~~~~~~~~~~~~~~~~~~~~
2nd method
Then #vec(PO)# will be perpendicular to #vec(P_1P_2)#
So #vec(OP_1)# is the projection of #vec(PP_1)# on #vec (P_2P_1#
Hence #absvec(OP_1)=vec(P_2P_1)*vec(PP_1)#
#=((-4hati+3hatk)*(-5hati-5hatk))/abs(-4hati+3k)#
#=(20-15)/sqrt((-4)^2+3^2)=1#
Again #vec(OP_2)# is the projection of #vec(PP_2)# on #vec (P_1P_2#
#absvec(OP_2)=vec(P_1P_2)*vec(PP_1)#
#=((4hati-3hatk)*(-hati-8hatk))/abs(4hati-3hatk)#
#=(-4+24)/sqrt(4^2+(-3)^2)=4#
So #vec(PO)# divides #P_1P_2# ar #O# in the ratio #OP_1:OP_2=1:4#
~~~~~~~~~~~~~~~~~~~~~~~~~~
Finally
#vec(PO)=(4*vec(PP_1)+1*vec(PP_2))/(1+4)#
#=>vec(PO)=(4(-5hati-5hatk)+1(-hati-8hatk))/5#
#=>vec(PO)=(-20hati-20hatk-hati-8hatk)/5#
#=>vec(PO)=(-21hati-28hatk)/5#
#=>abs(vec(PO))=1/5abs(-21hati-28hatk)#
#=>abs(vec(PO))=1/5*sqrt((-21)^2+(-28)^2)#
#=>abs(vec(PO))=1/5*sqrt(1225)=7# m
Time (t) taken by the bird to cover the distance #PO# is
#t="distance PO"/"velocity of the bird"=(7m)/(2m"/"s)=3.5# s
