A perpendicular is drawn from a point #P(1,6,3)# on the line #x/1=(y-1)/2=(z-2)/3#. Find the coordinates of the foot and length of perpendicular?

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Answer 1 · 2017-10-12T15:07:40

Coordinates of foot are #(1,3,5)# and length of perpendicular is #sqrt13#.

The equation of line is #x/1=(y-1)/2=(z-2)/3=t# and its direction ratios are #1,2,3# and a point on line is #Q(t,2t+1,3t+2)#.

Direction ratios of point #P(1,6,3)# and #Q# are

#t-1,2t+1-6,3t+2-3# or #t-1,2t-5,3t-1#

As it is perpendicular to the line, the dot product should be zero i.e.

#1(t-1)+2(2t-5)+3(3t-1)=0#

or #t-1+4t-10+9t-3=0#

or #14t=14# i.e. #t=1#

and coordinates of point #Q_0# are #(1,3,5)#

and #PQ_0=sqrt((1-1)^2+(3-6)^2+(5-3)^2)#

= #sqrt(0+9+4)=sqrt13#