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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