Hi!

I'm stuck on a vectors question on converting different forms of the same plane equation

Any help would be amazing!

I'm stuck on a vectors question on converting different forms of the same plane equation

Any help would be amazing!

It's going from

r=(3, 1,1)+a(1,3,-1)+b(2,5,0)

to r. (ci,d j, ek) = f

r=(3, 1,1)+a(1,3,-1)+b(2,5,0)

to r. (ci,d j, ek) = f

Original post by MayaVellichor

It's going from

r=(3, 1,1)+a(1,3,-1)+b(2,5,0)

to r. (ci,d j, ek) = f

r=(3, 1,1)+a(1,3,-1)+b(2,5,0)

to r. (ci,d j, ek) = f

What do you understand about the different forms?

Original post by mqb2766

What do you understand about the different forms?

I know that (ci, d j, ek) is the normal to the plane

And that r.n=a.n where r is general and a is a specific point on plane (I'm assuming that is the (3,1,1))

But I don't know how to find the normal from it

Original post by MayaVellichor

I know that (ci, d j, ek) is the normal to the plane

And that r.n=a.n where r is general and a is a specific point on plane (I'm assuming that is the (3,1,1))

But I don't know how to find the normal from it

And that r.n=a.n where r is general and a is a specific point on plane (I'm assuming that is the (3,1,1))

But I don't know how to find the normal from it

For the first form, it says that any point on the plane can be formed from a linear combination of the two direction vectors and adding that to the point you mention. So how are the two direction vectors and the plane normal related? A sketch may help (as usual).

(edited 10 months ago)

Original post by mqb2766

For the first form, it says that any point on the plane can be formed from a linear combination of the two direction vectors and adding that to the point you mention. So how are the two direction vectors and the plane normal related? A sketch may help (as usual).

are they directions along the plan and therefore the dot product of...

Ooooh the dot product of each vector with the normal will equal 0 cause of cos90? So then I solve simultaneously to find the normal?

Original post by MayaVellichor

are they directions along the plan and therefore the dot product of...

Ooooh the dot product of each vector with the normal will equal 0 cause of cos90? So then I solve simultaneously to find the normal?

Ooooh the dot product of each vector with the normal will equal 0 cause of cos90? So then I solve simultaneously to find the normal?

Yup, or use the cross product to get it directly (if youve covered it). Then once you have the normal, you can use the point to get d as per the previous post.

Original post by MayaVellichor

are they directions along the plan and therefore the dot product of...

Ooooh the dot product of each vector with the normal will equal 0 cause of cos90? So then I solve simultaneously to find the normal?

Ooooh the dot product of each vector with the normal will equal 0 cause of cos90? So then I solve simultaneously to find the normal?

Its also worth noting that even if you didnt understand the geometry insight, could sub equation 1 into equation 2. Then when you take the dot product with each termm, the left hand side would depend on a and b whereas the right hand side wouldnt. The only way this could occur for all a and b (free parameters) would be if the dot product of the corresponding direction vectors with the normal is zero in both cases.

While it falls out of the algebra, its "better" to understand the properties (geometry).

Original post by mqb2766

Its also worth noting that even if you didnt understand the geometry insight, could sub equation 1 into equation 2. Then when you take the dot product with each termm, the left hand side would depend on a and b whereas the right hand side wouldnt. The only way this could occur for all a and b (free parameters) would be if the dot product of the corresponding direction vectors with the normal is zero in both cases.

While it falls out of the algebra, its "better" to understand the properties (geometry).

While it falls out of the algebra, its "better" to understand the properties (geometry).

Yes I tried that where the normal is 0 but I don't entirely know what that means

Cause every plane should have a normal right?

Or is it more that they can take many values but they have to be in a certain ratio

Original post by MayaVellichor

Yes I tried that where the normal is 0 but I don't entirely know what that means

Cause every plane should have a normal right?

Cause every plane should have a normal right?

Not too sure what you mean, but if you have

n.(p + ar + bs) = d

where n, p, r and s are vectors as above, then the normal vector n must be non-zero and indeed its not unique, so any non-zero multiple of it is a normal vector as well. You can use the same representation of a line in 2d so

y = mx + c

-mx + y = c

(-m,1).(x,y) = c

where the normal is (-m,1) and it has a gradient -1/m as youd expect and d=c. Obv, a line is 1d in 2d space so a normal must exist, just as a plane is 2d in a 3d space, so again the a normal must exist. Its not unique as the same line is given by

(-2m,2).(x,y) = 2c

for instance. Its an equation, you can multiply through by a non-zero value.

So expanding you have

n.p + a n.r + b n.s = d

and the only way this can occur for all a and b is if n.r=n.s=0, so n is orhogonal to r and s. Then d = n.p

(edited 10 months ago)

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