1. ## Linear Algebra question

L has the parametric form x=s+1, y=s-2,z=2s+3 and P=3x-2y+z=1 and P'=x-y-2z=1. Find the points m on L and m' on L' such that the distance between them is minimised, this will occur when the line from m' to m is perpendicular to both L and L'
Would the L and L' be vectors or planes?
Thankyou

2. ## Re: Linear Algebra question

Originally Posted by grega
L has the parametric form x=s+1, y=s-2,z=2s+3 and P=3x-2y+z=1 and P'=x-y-2z=1. Find the points m on L and m' on L' such that the distance between them is minimised, this will occur when the line from m' to m is perpendicular to both L and L' Would the L and L' be vectors or planes?
There is much that is confusing about the way you have given this question.

In the question there is only one line & two planes given.
Yet the question asks about points on two lines, but there are not two lines.

You could find where the one line intersects each of the two planes.

3. ## Re: Linear Algebra question

Originally Posted by grega
L has the parametric form x=s+1, y=s-2,z=2s+3 and P=3x-2y+z=1 and P'=x-y-2z=1. Find the points m on L and m' on L' such that the distance between them is minimised, this will occur when the line from m' to m is perpendicular to both L and L'
Would the L and L' be vectors or planes?
Thankyou
Though it is not exactly a clear description, I believe that the line L' is the intersection of two planes P and P'. In such, solving P=0 and P'=0 together yields the parametric form of L' as follows.
$P - 2P' \implies x = -5z-1$ and $P - 3P' \implies y = -7z - 2$.
Therefore, parametric form of L' is $x=-5t-1$, $y=-7t-2$, and $z=t$.

The rest of the proof can be done using two methods as shown in the attached figure.

Note that L and L' are lines.