Thread: Matrices and vectors point-line distance

I can't find the shortest distance from the point P=<-5,2,-6> to a point on the line given by l:<x,y,z>=<1t,3t,6t>. Can someone help? please?

This formula is for R^2 but I need R^3
Should I use it as ...cz+d=0 and ...b^2+c^2? If so how.

SillyMath

I understand that the parametric equations of the line given to you are x=1t , y=3t ,z=6t
therefore the vector v parallel to this line is v=(1t,3t,6t) expressed as function of t.
Now get one point of the line Q(x,y,z)=Q(1t,3t,6t) and define the vector PQ=(t+5,3t-2,6t+6)

Now get the dot product of the vector v x PQ =0 and determine the value of t.
thus you have the value of t and you can find the exact coordinates of the point Q and of course the coordinates of the vector PQ.
GET THE |PQ| AND THIS IS THE DISTANCE OF THE POINT P FROM THE LINE GIVEN TO YOU.

MINOAS

Originally Posted by SillyMath

I can't find the shortest distance from the point P=<-5,2,-6> to a point on the line given by l:<x,y,z>=<1t,3t,6t>. Can someone help? please?

Given the line and a point the distance from the point to the line is

.

The shortest distance is always measured "perpendicularly".

The line given is x= t, y= 3t, z= 6t. Any plane perpendicular to the line must be of the form . Taking , the plane perpendicular to the line and containing (-5, 2, -6) is (x+ 5)+ 3(y- 2)+ 6(z+ 6)= x+3y+ 6z+ 35= 0.
The line x= t, y= 3t, z= 6t crosses that plane when t+ 3(3t)+ 6(6t)+ 35= 46t+ 35= 0.

Solve that for t and solve for x, y, and z where the given line crosses that plane. The distance from that point to (-5, 2, -6) is the shortest distance from the given line to the given point.