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**0001T****Member**- Registered: 2007-03-28
- Posts: 4

Hi, I just got to know about this forum & I got this little problem...

I am trying to solve this cross product; 3 vertices p1=(0,0,3), p2=(4,0,0) & p3=(0,4,3) given, I need to find the normal of the surface bounded by these 3 vertices.

I was told answer is (3,0,4) but my attempts kept leading me to (12,0,16)

here's what I did.

normal = (p1-p2) x (p3-p2) = [(0,0,3)-(4,0,0)]x[(4,0,0)-(0,4,3)] = (-4,0,3)X(4,-4,-3)

normal = (12,0,16)

where did i go wrong?

Thank you

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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

I dont think you did wrong , cuz 4(3,0,4)=(12,0,16)

Numbers are the essence of the Universe

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**0001T****Member**- Registered: 2007-03-28
- Posts: 4

Stanley_Marsh wrote:

I dont think you did wrong , cuz 4(3,0,4)=(12,0,16)

so if i present the surface normal as (12,0,16) it is not considered as wrong?

thanks

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

What about a unit-length normal vector??

3^2 + 4^2 = 5^2, so divide by 5.

Unit vector normal: (3/5, 0, 4/5)

Why are you so concerned about the magnitude of the normal.

Just the angle being right is good enough, isn't it?

**igloo** **myrtilles** **fourmis**

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**0001T****Member**- Registered: 2007-03-28
- Posts: 4

I see.

Thank you so much.

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**0001T****Member**- Registered: 2007-03-28
- Posts: 4

this is a bit embarassing...because i had forgotten the basics:(

how do i solve this

-5db = 20log M(f)

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