Question #26a9b

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`

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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