Distances between planes (intro lin alg class)

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Homework Help Overview

The discussion revolves around finding the distance between two planes defined by the equations x + y - 2z = 0 and 3x + 3y - 6z = 1, within the context of linear algebra and vector analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for calculating the distance between the planes and the selection of appropriate normal vectors. Questions arise regarding the choice of normal vector and whether it matters which plane is used to extract it.

Discussion Status

Participants are actively engaging with the problem, clarifying the use of normal vectors and exploring the implications of parallel planes. Some guidance has been provided regarding the flexibility in choosing normal vectors, but no consensus on the final answer has been reached.

Contextual Notes

There is a focus on understanding the relationship between the normal vectors of the planes and their implications for distance calculations. Participants are also navigating potential discrepancies in their results.

Square1
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Homework Statement



Im working currently with vectors. The question asks for the distance between two planes given by the two following equations:
x + y -2z = 0
3x + 3y -6z = 1

Homework Equations



I know the planes, H1, and H2 are parallel, so I can pick any random point on either, call the point Q, use the other plane as my H, and apply the following formula:

distance(H,Q) = (q-p) dot (n/|n|)



The Attempt at a Solution



Im assuming that someone familiar with vectors and lin. alg is going to answer this so I am not writing the whole essay. But yea if more explanation is needed then let me know.

For H I chose the first plane, so that n from plane 1 is (1, 1, -2). Correct me if I am wrong but I understand that you do need to pick the appropriate plane/line to use the right n? Or can it be either?
For Q on plane 2 I did - (1/3, 0, 0)
For a point P on plane 1 I chose (0,0,0)

After using the corresponding position vectors for the two points, I plug them in the formula and get the result:

(1)/(3√6), but my answers say it is (√6)/(9)

Thanks all.
 
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Square1 said:
distance(H,Q) = (q-p) dot (n/|n|)

Correct

Square1 said:
For H I chose the first plane, so that n from plane 1 is (1, 1, -2).

Correct

Square1 said:
For Q on plane 2 I did - (1/3, 0, 0)
For a point P on plane 1 I chose (0,0,0)
You mean (1/3, 0,0,) for Q, do you?

Square1 said:
After using the corresponding position vectors for the two points, I plug them in the formula and get the result:

(1)/(3√6), but my answers say it is (√6)/(9)

Thanks all.

The correct answer is 1/(3√6)

ehild
 
ehild said:
You mean (1/3, 0,0,) for Q, do you?

Yea. I use (1/3, 0, 0) as my Q, and purposefully taking it from plane 2.

Would you mind following up on the other question I snuck in there as well? About picking correct "n"?
 
Square1 said:
Would you mind following up on the other question I snuck in there as well? About picking correct "n"?

Yes, of course. Show the problem.

ehild
 
Square1 said:
For H I chose the first plane, so that n from plane 1 is (1, 1, -2). Correct me if I am wrong but I understand that you do need to pick the appropriate plane/line to use the right n? Or can it be either?

Ok so that is where the question originated but I will rephrase it...

When picking "n", if the lines are parallel, I can pick either plane 1 or plane 2 to extract "n" right? Or is there any sort of decision I need to make?

Furthermore, if I were to pick the second plane 3x + 3y -6z = 1, my "n" from this plane could be any multiple of that expression, which is why I could say n = (1,1,-2) or n = (3, 3, -6).

Thanks.
 
The equation of a plane comes from that any vector in the plane is perpendicular to the normal vector, so their dot product is zero: (r-p)n=0. In components: xnx+yny+znz=pxnx+pyny+pznz=constant

The magnitude of the normal vector does not influence the zero dot product. If it is (1,1,-2) , you equally can choose any vector which is a multiple of this. So (3,3,-6) is also a normal vector, or (-1,-1,2)...Often we choose a unit normal vector, with the components divided by the magnitude.

In the problem, you had two planes. And you found that the normal vectors were multiples of each other. If the normal vectors are of identical direction, the planes are parallel.

ehild
 
Thanks so much!
 
Square1 said:
Thanks so much!

You are welcome:smile:

ehild
 

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