# How do you work out the normal of a plane

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#1
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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1 year ago
#2
(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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#3
(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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1 year ago
#4
(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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#5
(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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1 year ago
#6
(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.

You're aware of the dot product though, right?
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#7
(Original post by RDKGames)

You're aware of the dot product though, right?
Yep. I know the dot product
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#8
(Original post by MM2002)
Yep. I know the dot product
Is there any point learning the cross product?!
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1 year ago
#9
(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 such that it is perpendicular to both and . So two conditions must be satisfied:

and

These are two equations in three variables. You have one extra degree of freedom. So you can set and solve for to get the normal vector.
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1 year ago
#10
(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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