Points on lines with parametric equations (linear algebra)

[fattycakez]

Homework Statement

"Let L1 be the line having parametric equations : x = 2 - s, y = -1 + 2s, z = 1+s and L2 be the line:
x = 1 +t, y = 2+ t, z =2t .

a. Do the lines intersect? If so, find the point of intersection.

b. Find the point P on the graph of L1 that is closest to the graph of L2 and find the point Q on the graph of L2 that is closest to the graph of L1. Hint: Use the fact that the vector PQ will be orthogonal to the direction vectors of both lines. "

Homework Equations

The Attempt at a Solution

In part a, I set the parametric equations equal to each other and solved for t and s. It looks like the lines do not intersect.

I'm not sure how to go about part b. How does the hint that the vector PQ will be orthogonal to the direction vectors help me?
The direction vectors would be:
L1 = (-1, 2, 1) L2 = (1, 1, 2)

Any help is greatly appreciated!

 

[Mark44]

Homework Statement

"Let L1 be the line having parametric equations : x = 2 - s, y = -1 + 2s, z = 1+s and L2 be the line:
x = 1 +2, y = 2+ t, z =2t .

Should the x-coordinate of L2 be x = 1 + 2t?

a. Do the lines intersect? If so, find the point of intersection.

b. Find the point P on the graph of L1 that is closest to the graph of L2 and find the point Q on the graph of L2 that is closest to the graph of L1. Hint: Use the fact that the vector PQ will be orthogonal to the direction vectors of both lines. "

Homework Equations

The Attempt at a Solution

In part a, I set the parametric equations equal to each other and solved for t and s. It looks like the lines do not intersect.

I'm not sure how to go about part b. How does the hint that the vector PQ will be orthogonal to the direction vectors help me?
The direction vectors would be:
L1 = (-1, 2, 1) L2 = (1, 1, 2)

Any help is greatly appreciated!

 

[fattycakez]

Ahh sorry, It should be x = 1+ t

 

[Mark44]

I'm not sure how to go about part b. How does the hint that the vector PQ will be orthogonal to the direction vectors help me?
The direction vectors would be:
L1 = (-1, 2, 1) L2 = (1, 1, 2)

At the points on the lines that are closest to each other, the segment joining the two lines will be perpendicular to each line.

 

[fattycakez]

Okay, will it have something to do with PQ ⋅ L1 = 0 and PQ ⋅L2 = 0? Or am I way off here?

 

[LCKurtz]

Mark's hint is different from the hint given in the original post. Take a general point on each line, one depends on $s$ and the other on $t$. Write the distance (easier to use distance^2) between those two points. Minimize that using calculus methods to minimize a function of two variables. The $s$ and $t$ you get will give you the two points you seek.

 

[fattycakez]

What would a general point on the parametric lines look like? The x, y, z components give in the problem?
( I haven't taken multi variable calc yet and this class assumes I will only use algebra to complete the homework)

 

[Mark44]

Mark's hint is different from the hint given in the original post.

I don't see how what I said was different. In the OP, it states that "Hint: Use the fact that the vector PQ will be orthogonal to the direction vectors of both lines."

Take a general point on each line, one depends on $s$ and the other on $t$. Write the distance (easier to use distance^2) between those two points. Minimize that using calculus methods to minimize a function of two variables. The $s$ and $t$ you get will give you the two points you seek.

A completely different approach uses the hint above by taking the cross product of the two direction vectors. Then, take two arbitrary points, one from each line, and create a displacement vector between those two points. The shortest distance between the two lines will have to be a scalar multiple of the vector you obtained from the cross product.