How do you work out the normal of a plane

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MM2002
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Lets say we have the plane equation: 3,5,-1 + lambda -1,-7,5 +mu 1,-2,-3 . How would I work out the normal to this plane?
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RDKGames
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(Original post by MM2002)
Lets say we have the plane equation: 3,5,-1 + lambda -1,-7,5 +mu 1,-2,-3 . How would I work out the normal to this plane?
The plane is defined by a linear combination of the two vectors (-1,-7,5) and (1,-2,-3). Hence these vectors lie entirely in the plane. Any vector perpendicular to the plane must be perpendicular to both of these vectors.

So... cross product ?
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MM2002
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(Original post by RDKGames)
The plane is defined by a linear combination of the two vectors (-1,-7,5) and (1,-2,-3). Hence these vectors lie entirely in the plane. Any vector perpendicular to the plane must be perpendicular to both of these vectors.

So... cross product ?
Cross product if the lambda vector and mu vector. But, you just get a singular value . I think I phrased my question poorly. the normal to that plane is (1,2,3) but how did they show that.
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RDKGames
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(Original post by MM2002)
Cross product if the lambda vector and mu vector. But, you just get a singular value . I think I phrased my question poorly. the normal to that plane is (1,2,3) but how did they show that.
Cross product between two vectors is another vector ...

Dot product between two vectors is a scalar.

Don't get the two confused.

And no the normal vector to that plane is not (1,2,3). It's any multiple of the vector (31,2,9).
https://www.wolframalpha.com/input/?...1%2C-2%2C-3%29
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MM2002
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(Original post by RDKGames)
Cross product between two vectors is another vector ...

Dot product between two vectors is a scalar.

Don't get the two confused.

And no the normal vector to that plane is not (1,2,3). It's any multiple of the vector (31,2,9).
https://www.wolframalpha.com/input/?...1%2C-2%2C-3%29
Right. Haven't been taught that yet. But, the solution bank says (1,2,3) so, I will assume its just wrong.
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RDKGames
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(Original post by MM2002)
Right. Haven't been taught that yet. But, the solution bank says (1,2,3) so, I will assume its just wrong.
Maybe it was taken out of spec, think I heard something about this.

You're aware of the dot product though, right?
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MM2002
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(Original post by RDKGames)
Maybe it was taken out of spec, think I heard something about this.

You're aware of the dot product though, right?
Yep. I know the dot product
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MM2002
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(Original post by MM2002)
Yep. I know the dot product
Is there any point learning the cross product?!
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RDKGames
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(Original post by MM2002)
Yep. I know the dot product
Ok, so you are probably aware that the dot product between two perpendicular vectors is zero.

So you seek a normal vector \mathbf{n} = (n_1,n_2,n_3) such that it is perpendicular to both (-1,-7,5) and (1,-2,-3). So two conditions must be satisfied:

(n_1,n_2,n_3) \cdot (-1,-7,5) = 0
and
(n_1,n_2,n_3) \cdot (1,-2,-3) = 0


These are two equations in three variables. You have one extra degree of freedom. So you can set n_2 = 2 and solve for n_1,n_3 to get the normal vector.
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RDKGames
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(Original post by MM2002)
Is there any point learning the cross product?!
Sure, if you want to be able to quickly calculate the perpendicular vector to two other vectors.
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