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## Math question

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### #1DmitryNik  Members

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Posted 04 November 2012 - 07:38 AM

Hello.

Recently I found an interesting assignment. I did it almost 2 years ago and some how I figured out how to do it then. But not now. Could anybody, please, explain me, why this particular problem was solved with a help of the dot product? See problem below.

We have a line l=[0, 1, 2] + a*[-3, 4, 7], which is perpendicular to the plane W. Also plane W contains the point Q=[3, 5, -3]. We have to find the equation of the plane W.

The solution was: const = dot_product([-3, 4, 7], [3,5,-3]) => W's equation is -3x + 4y + 7z = const.

Problem is, I can't remember, what I was thinking of back to those days. I need to understand this solution for solving one problem in the game.
I'll repeat the question here once again:

Could anybody, please, explain me, why this particular problem was solved with a help of the dot product?

### #2SiCrane  Moderators

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Posted 04 November 2012 - 08:46 AM

If -3x + 4y + 7z = const is the equation for the plane, then this equation must be true for every point on the plane. Q is a point on the plane so -3*3 + 4*5 + 7*-3 = const must be true. -3*3 + 4*5 + 7*-3 is the same as (-3, 4, 7)·(3, 5, -3).

### #3DmitryNik  Members

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Posted 04 November 2012 - 08:53 AM

If -3x + 4y + 7z = const is the equation for the plane, then this equation must be true for every point on the plane. Q is a point on the plane so -3*3 + 4*5 + 7*-3 = const must be true. -3*3 + 4*5 + 7*-3 is the same as (-3, 4, 7)·(3, 5, -3).

Thank you. You helped me. I guess, I have to read my old notes through more carefully =)

### #4incertia  Members

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Posted 04 November 2012 - 09:22 AM

A normal vector is perpendicular to all vectors in a plane. Now let be a point on the plane. We can now form a vector on the plane with . For this vector to be perpendicular to the normal vector (given by the line), their dot product must equal 0. So we have:

Distributing the dot product yields

And expanding yields

what

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