# Thread: tough vector problem please help

1. ## tough vector problem please help

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).

2. Originally Posted by myoplex11
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

3. i still don't get it.

4. Here is another way to do it.
Write the equation of the plane through the point that is perpendicular to the line.
Then find the point on the line at which the plane intersects the line.

5. 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.

6. You have the correct idea. However, I did not check the algebra or arithmetic.

7. 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: .$\displaystyle P\left(\text{-}\frac{1}{5},\:\frac{8}{5},\:1\right)$