How can I find the shortest distance between the point `(0,1,-1)` and the line `(x,y,z) = (2,1,3) + t(3,-1,-2)`?

Let `p_0=(0,1,-1)` and
`l->p=p_1+t vec v`

with `p=(x,y,z)`, `p_1=(2,1,3)` and `vec v = (3,-1,-2)`

The square distance between `p` and `p_0` is given by

`d^2 = norm(p-p_0)^2`

substituting `p` we have

`d^2=norm(p_1+t vec v-p_0)^2`.

Developing we have

`d^2=norm(p_1-p_0)^2+2 t << p_1-p_0, vec v >> + t^2 norm (vec v)^2`

Here `<< cdot, cdot >>` represents the scalar product of two vectors.

The minimum distance from `p_0` to `l` is attained at `t_0` such that `d^2` has minimum value.

This occurs at a point `t_0` such that

`d/(dt)d^2=2<< p_1-p_0, vec v >> +2t_0 norm(vec v)^2=0`

or

`t_0 = -(<< p_1-p_0, vec v >>)/norm(vec v)^2` and the shortest distance is given by

`d = sqrt(norm(p_1-p_0)^2+2 t_0 << p_1-p_0, vec v >> + t_0^2 norm (vec v)^2)`

or simplifying

`d = sqrt(norm(p_1-p_0)^2-((<< p_1-p_0, vec v >>)/norm(vec v))^2)`

Here `(<< p_1-p_0, vec v >>)/norm(vec v)` is the projection of the segment `p_1-p_0` into the line direction (`vec v`) so it is an interesting form in which the Pythagoras theorem appears.

Finally, substituting values we obtain.

`d = sqrt(138/7) approx 4.44008`

We have line:

`vec l(t) = langle 2,1,3 rangle + t langle 3, -1 ,-2rangle`

and point `P = (0,1,-1)`

Let `Q` be the point on line `vec l` that represents the shortest distance between `P` and `vec l`, with `Q = vec l(alpha)`, so:

`Q = ( 2+ 3 alpha, 1 - alpha,3 - 2 alpha )`

Now, `vec(PQ)` is the direction of the line of shortest distance going from `P" to "Q`, and is:

`vec(PQ) = langle -(0,1,-1) + ( 2+ 3 alpha, 1 - alpha,3 - 2 alpha ) rangle = langle 2 + 3 alpha, - alpha, 4 - 2 alpha rangle qquad triangle`

And `vec(PQ)` is going to be perpendicular to `vec (l)_d` (where `vec l_d = langle 3, -1 ,-2rangle` is the direction vector of line `vec l)`, this being a fact of the geometry and the trade-off for not using calculus.

Therefore:

`vec(PQ) cdot vec l_d = 0`.

`implies langle 2 + 3 alpha, - alpha, 4 - 2 alpha rangle cdot langle 3, -1 ,-2 rangle = 0`

`implies alpha = 1/7 [implies Q = ( 17/7, 6/7,19/7 )]`

From `triangle` above:

`abs( vec(PQ)) = sqrt( ( 2 + 3 alpha)^2 + (-alpha)^2 +( 4 - 2 alpha)^2)`

`= sqrt(138/7) approx 4.4 " units"`

The direction of the given line is the vector `( 3,-1,-2 )`

Because the above vector must be a normal vector to the plane that contains the point closest to the given point, we know that its general form is:

`3x -y-2z = c`

Find the specific plane that contains the given point, `(0,1,-1)`, by substituting and then solving for c:

`3(0) -1-2(-1) = c`

`c = 1`

The plane that contains the nearest point to the point `(0,1,-1)` is:

`3x-y-2z=1" [1]"`

The parametric equations of the line are:

`x = 3t+2`
`y = 1 -t`
`z = 3 - 2t`

Substitute these equations into equation [1] and solve for t:

`3(3t+2)-(1 -t)-2(3 - 2t)=1`

`9t+6 +t -1+4t-6 = 1`

`14t - 1=1`

`14t=2`

`t = 2/14 = 1/7`

Using the parametric equations and the value of t, we can find the nearest point:

`x = 3(1/7)+2`
`y = 1 -1/7`
`z = 3 - 2(1/7)`

`x = 3/7+14/7`
`y = 7/7 -1/7`
`z = 21/7 - 2/7`

`x = 17/7`
`y = 6/7`
`z = 19/7`

Use the distance formula to find the distance from, `(0,1,-1)` to `(17/7,6/7,19/7)`

`d = sqrt((17/7-0)^2+(6/7-1)^2+(19/7--1)^2)`

`d = sqrt((17/7)^2+(-1/7)^2+(26/7)^2)`

`d = sqrt(966)/7`

`d ~~ 4.44`