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

3 Answers
Jul 7, 2016

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

Jul 7, 2016

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]

Jul 7, 2016

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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