Question

Show that the perpendicular distance of a point a from the line r=b+tc is

Answer 1

Answer:

s

order by

up vote

4

down vote

accepted

Hint: A point on the line BC is described by tB+(1−t)CtB+(1−t)C, t∈Rt∈R. Try to minimize its distance to AA

share improve this answer

answered

Aug 27 '16 at 19:11

H. H. Rugh

23.9k●11 gold badge●1212 silver badges●3636 bronze badges

up vote

12

down vote

Intuitively, you want the distance between the point A and the point on the line BC that is closest to A. And the point on the line that you are looking for is exactly the projection of A on the line. The projection can be computed using the dot product (which is sometimes referred to as "projection product").

So you can compute the direction vector dd of the line BCBC. This is the difference of BB and CC, divided by their distance:

d=(C−B)/||C−B||

d=(C−B)/||C−B||

Then you can define a vector from BB to AA:

v=A−B

v=A−B

Computing the dot product between this vector and the direction vector will give you the the distance between BB and the projection of AA on BCBC:

t=v⋅d

t=v⋅d

The actual projection PP of AA on BCBC is then given as

P=B+t⋅d

P=B+t⋅d

And finally, the distance that you have been looking for is

||P−A||