Let L be the line that passes through the points (1,1,1) and(3,0,1). Find the point on the line L that is closest to the point (1,4,5).
The line that passes through these points may be written:
$\displaystyle L(\lambda)=(1,1,1)+\lambda [(3,0,1)-(1,1,1)] = (1,1,1)+\lambda (2,-1,0) = (1+2\lambda, 1-\lambda, 1)$
So the problem is now to find $\displaystyle \lambda$ that minimises the distance from $\displaystyle L(\lambda)$ to $\displaystyle (1,4,5)$
(note it is usualy easier to minimise the square distance rather than the distance itself)
RonL
thanks. is this correct?
v=(-2,1,0) is a vector parallel to the line so it a normal vector to plane containing the point(1,4,5).so using the components of the normal vector and the point(1,4,5) i got equation of the plane as -2x+y-2=0. The parametric equations of the line i got as x=3-2t, y=t, z=1. substituting these values of x,y and z into the plane equation and solving for t=7/5. plugging this value of t back into the parametric equations gives the point.
Hello, myoplex11!
I have a primitive (non-vector, non-calculus) solution . . .
Let $\displaystyle L$ be the line that passes through the points $\displaystyle (1,1,1)$ and $\displaystyle (3,0,1)$.
Find the point $\displaystyle P$ on the line $\displaystyle L$ that is closest to the point $\displaystyle Q(1,4,5)$
The direction vector of $\displaystyle L$ is: .$\displaystyle \langle2,\,-1,\,0\rangle$
The parametric equations of line $\displaystyle L$ are: .$\displaystyle \begin{Bmatrix}x & = & 1 + 2t \\ y & = & 1 - t \\ z & = & 1 \end{Bmatrix}$
Let $\displaystyle P$ be any point on line $\displaystyle L\!:\;P(1+2t,\:1-t,\:1)$
Consider $\displaystyle d \:=\:\overline{PQ}$
We will minimize: $\displaystyle D \:=\:d^2\:=\:[(1+2t) -1]^2 + [(1-t) - 4]^2 + [1- 5]^2$
Simplify: .$\displaystyle D \:=\:5t^2 + 6t + 25$
This is an up-opening parabola; its minimum occurs at its vertex.
Vertex formula: .$\displaystyle v \:=\:\frac{\text{-}b}{2a}$
We have: .$\displaystyle a = 5,\:b = 6$
Hence, the vertex is at: .$\displaystyle t \:=\:\frac{\text{-}6}{2(5)} \:=\:\text{-}\frac{3}{5}$
Then: .$\displaystyle \begin{array}{ccccc}x & = & 1 + 2\left(\text{-}\frac{3}{5}\right) & = & \text{-}\frac{1}{5} \\ y & = & 1 - \left(\text{-}\frac{3}{5}\right) & = & \frac{8}{5} \\ z & = & 1 & = & 1 \end{array} $
Therefore: .$\displaystyle P\left(\text{-}\frac{1}{5},\:\frac{8}{5},\:1\right)$