Question

How to find the distance from the point A(3,-5,5) to the line x = 2 + 3t , y= 1-2t, z= -1 + t ?

Answer 1

`5/sqrt6`

Explanation:

there is a equation
`x+2y+z-3=0`
use distance formula
=`((1*3-5*2+5*1)-3)/sqrt(1^2+2^2+1^2)`
=`-5/sqrt6`

`abs(-5/sqrt6)`
=`5/sqrt6`

Answer 2

`sqrt[83/2]`

Explanation:

Defining

`p_0 = {2,1,-1}`
`vec v = {3,-2,1}`
`p_A={3,-5,5}`

we have to determine the distance between the line
`r->p_0+t vec v` and the point `p_A`

Using Pitagoras we have

`a = norm(p_a-p_0)`

`b = abs(<< p_A-p_0,(vec v)/norm(vec v)>>)`

`d = sqrt(a^2-b^2)` which is the sought distance

`a = sqrt((3-2)^2+(-5-1)^2+(5+1)^2`

`(vec v)/norm(vec v) =( {3,-2,1})/sqrt(3^2+2^2+1)`

`b = abs(((3-2)cdot 3+(5+1)cdot 2+(5+1)cdot 1)/sqrt(3^2+2^2+1))`

Finally

`d = sqrt[83/2]`

Answer 3

`sqrt(83/2).`

Explanation:

We find the co-ords. of the foot `M` of the perp. from `A(3,-5,5)` on the given line `L : x=2+3t,y=1-2t,z=-1+t, t in RR.`

We take a note that since `M in L, M(2+3t,1-2t,-1+t)` for some `t in RR.`

Also `A(3,-5,5) rArr vec(AM)=(2+3t-3,1-2t+5,-1+t-5)=(3t-1,6-2t,t-6)`

The Direction Vector `vecl` of line `L` is `vecl=(3,-2,1)`

Knowing that `vec(AM)` is perp. to `vecl`, we have, `vec(AM).vecl=0 rArr (3t-1,6-2t,t-6).(3,-2,1)=0`
`:. 3(3t-1)-2(6-2t)+(t-6)=0`
`:. 9t-3-12+4t+t-6=0`
`:. 14t=21 rArr t=3/2 rArr vec(AM)=(9/2-1,6-3,3/2-6)=(7/2,3,-9/2)`

Hence the Dist. `AM=||vec(AM)||=sqrt{49/4+9+81/4)=sqrt(166/4)=sqrt(83/2),` as derived by Cesareo R. Sir!
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