# Calculating Distance Between Parallel Lines in 3D Space

• Dr Zoidburg
In summary,The lines P and Q in 3D are parallel, but the distance between them is not perpendicular to the lines.
Dr Zoidburg

## Homework Statement

following equations represent lines P nad Q in 3 dimensional space:
P: 2-x=y+1=(z-3)/2
Q: 5-x=y-2=(z+1)/2
(a) Show that P & Q are two different lines which are parallel
(b) Find the distance between the two lines, measured along a line perpendicular to P & Q.

## The Attempt at a Solution

(a) I attempted to turn the above equations into vector equations. I came out with the following:
v(P) = (2,-1,3) + t(-1,1,2)
v(Q) = (5,2,-1) + s(-1,1,2)
I concluded that since the the direction vectors are the same, therefore the lines are parallel.
correct?
(b) Not sure what to do here. Do I use the distance formula for between two points?
ie.
$$\sqrt{(2-5)^{2}+(-1-2)^{2}+(3--1)^{2}}$$
I then get $$\sqrt{34}$$
correct? (I have the nagging feeling it ain't!)

Last edited:
Dr Zoidburg said:

## Homework Statement

following equations represent lines P nad Q in 3 dimensional space:
P: 2-x=y+1=(z-3)/2
Q: 5-x=y-2=(z+1)/2
(a) Show that P & Q are two different lines which are parallel
(b) Find the distance between the two lines, measured along a line perpendicular to P & Q.

## The Attempt at a Solution

(a) I attempted to turn the above equations into vector equations. I came out with the following:
v(P) = (2,-1,3) + t(-1,1,2)
v(Q) = (5,2,-1) + s(-1,1,2)
I concluded that since the the direction vectors are the same, therefore the lines are parallel.
correct?
Yes, the vector equations and the conclusion that the lines are parallel because they have the same direction vector are correct.

(b) Not sure what to do here. Do I use the distance formula for between two points?
ie.
$$\sqrt{((2-5)}^{2}+(-1-2)}^{2}+(3--1)}^{2})$$
I then get $$\sqrt{34}$$
correct? (I have the nagging feeling it ain't!)
Okay, but what two points? Do you have any reason to believe that the line from (2, -1, 3), on the first line, to (5, 2, -1), on the second line, is perpendicular to the lines? One way to check is to calculate the vector from one of those points to the other:(5-2, 2-(-1), -1-3)= (3, 3, -4) and then find its dot product with the direction vector: (3, 3, -4)$\cdot$(-1, 1, 2)= -3+ 3- 8= -8. Since that is NOT 0, the two vectors are NOT perpendicular and the line between the two points is NOT perpendicular to the two lines.

Try this instead: the vector (-1, 1, 2), the direction vector, is in the direction of the line and so perpendicular to any plane perpendicular to the line. That means that the plane perpendicular to the first line and containing the point (2, -1, 3) has equation -1(x- 2)+ 1(y-(-1))+ 2(z- 3)= 0 or -x+ y+ 2z= 3. Where does the second line x= 5- s, y= 2+ s, z= -1+ 2s cross that line? (Replace x, y and z in the equation of the plane with those formulas in terms of s and solve for s.) Now you can find the distance between (2, -1, 3) and that point.

ah-ha! I knew I was missing a step there. It seemed too easy, just using the two points provided but I had a massive brainfreeze as to why this wasn't correct.

So in the equation you supplied (TQ very much!):
-x+ y+ 2z= 3,
we sub
x= 5- s
y= 2+ s
z= -1+ 2s
and solve for s:

-(5-s) + (2+s) + 2(-1+2s) = 3
-5+2-2 + s + s + 4s = 3
6s=6
s=1

point on Q which is perpendicular to P at (2,-1,3) is (5,2,-1) + 1(-1,1,2) = (4,3,1).
Then use distance formula:
$$\sqrt{(2-4)^{2}+(-1-3)^{2}+(3-1)^{2}}$$
=$$\sqrt{24}$$

yes/no?

Last edited:
Well, I was under the impression that -1- 2 was equal to -3, not 1!

D'oh!

okay.got it now. thanks for the help.

## 1. What is a vector problem?

A vector problem is a type of mathematical problem that involves using vectors, which are quantities that have both magnitude and direction, to solve it. These types of problems often involve finding the magnitude and direction of a vector, or performing operations such as addition, subtraction, and multiplication on vectors.

## 2. How do I solve a vector problem?

To solve a vector problem, you will need to have a good understanding of vector properties and operations. First, identify the given information and what you are being asked to find. Then, use vector addition, subtraction, and/or multiplication to manipulate the vectors and find the desired result. Don't forget to consider the direction of the vectors as well as their magnitudes.

## 3. What are some real-life applications of vector problems?

Vector problems have many real-life applications in fields such as physics, engineering, and navigation. For example, calculating the velocity and acceleration of an object, determining the forces acting on a structure, and finding the shortest route between two points are all examples of problems that can be solved using vectors.

## 4. What is the difference between a scalar and a vector?

A scalar is a quantity that has magnitude but no direction, while a vector has both magnitude and direction. For example, speed is a scalar quantity because it only tells us how fast an object is moving, while velocity is a vector quantity because it tells us both the speed and direction of the object's motion.

## 5. What are some common mistakes when solving vector problems?

One common mistake when solving vector problems is forgetting to consider the direction of the vectors. Another mistake is not using the correct vector operations, such as adding or subtracting vectors component-wise instead of using the vector sum or difference formula. It's also important to be careful with units and make sure they are consistent throughout the problem.

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