Lines and planes in space

[tjkubo]

I don't think this fits into calculus very much but here goes.
How do you find the point on a line closest to a given point in space?
For example how would you solve this problem:
Find the coordinates of point P on the line X = <0, 1, 0> + t<1, 0, 3> nearest to the point A(1, 2, 3).

 

[Opalg]

The point P has to lie in the plane through A perpendicular to the line X. The line has direction <1, 0, 3>, so the plane has equation x + 3z = const. The fact that the plane contains A will tell you the value of the constant. Then you can find the value of t such that the point <0, 1, 0> + t<1, 0, 3> lies in the plane. That will be the point on X that is closest to A.

 

[Deleted member 4993]

I don't think this fits into calculus very much but here goes.
How do you find the point on a line closest to a given point in space?
For example how would you solve this problem:
Find the coordinates of point P on the line X = <0, 1, 0> + t<1, 0, 3> nearest to the point A(1, 2, 3).

Another way:

Find the equation of the line normal to X and passing through P. Let that line be PQ.

Now find the point of intersection between PQ and X.

Incidentally, there is a standard formula for this - you can find it in a mathematical handbook (Schaum series) or through Google.

 

[soroban]

Hello, tjkubo!

This can be solved without Calculus . . .

Find the coordinates of point \(\displaystyle P\) on the line: \(\displaystyle \langle0,\,1,\,0\rangle\,+\,t\langle1,\,0,\,3\rangle\)
nearest to the point \(\displaystyle A(1,\,2,\,3)\).

We have the point \(\displaystyle P:\;\begin{array}{ccc}x & = & t \\ y & = & 1 \\ z & = & 3t\end{array}\;\) and point \(\displaystyle A1,\,2,\,3)\)

The distance between them is: \(\displaystyle \:d \:=\:\sqrt{(t-1)^2\,+\,(1-2)^2\,+\,(3t-3)^2}\)

Let: \(\displaystyle \,D\;=\;d^2\;=\;(t-1)^2\,+\,(-1)^2\,+\,(3t-3)^2\)

. . . .\(\displaystyle D\;= \;10t^2\,-\,20t\,+\,11\) . . . and we will minimize \(\displaystyle D\).


This is an up-opening parabola; its minimum value occurs at its vertex.

Since the vertex is at: \(\displaystyle \,t\:=\:\frac{-b}{2a}\), and we have: \(\displaystyle \,a\,=\,10,\;b\,=\,-20,\;c\,=\,11\)

. . then: \(\displaystyle \:t \:=\:\frac{-(-20)}{2(10)} \:=\:1\)


Therefore, point \(\displaystyle P \;=\;(1,\,1,\,3)\)