Planes perpendicular to vectors

In summary, the equation of the norm of the plane is influenced by a and b. For b, I think the dot product could be used if I set it to zero.
  • #1
riskybeats
18
0
a)Find the equation of the tangent plane to the graph z = x2 + 4y2 at the point (a, b, a2 + 4b2)

b) For what values of a and b is the tangent plane perpendicular to the vector (3, -4, 2)?

I figured you have to use the tangent plane equation z - zo = zx(a,b)(x-a) + zy(a,b)(y-b). This gave me the equation

Z = 2ax + 8by - a2 - 4b2

For b, I think the dot product could be used if I set it to zero. But how would I go about solving for a and b? I think I may just be getting tired and a bit sloppy, because I know this is already stuff I have gone over, the only new thing is solving for a and b.

Any direction on this would be great, I'm just having trouble visualizing this problem.
 
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  • #2
I couldn't get to sleep because I kept on thinking this problem through. As far as I can see, if the plane's norm is the same as the vector <3, -4, 2>, then it must be perpendicular to it. I am not sure how to make the norm on the plane into that though based off of just of values of a and b. Does anyone have any suggestions?
 
  • #3
Doesn't have to be the same, just parallel, i.e. costheta=1 or -1.

Find the equation of the norm of the plane first.
 
  • #4
The norm of my plane is influenced by a and b. This is what I have:

2a (X - a) + 8b (y - b) - z + a^2 + 4b^2 = 0

So if the vector is <3, -4, 2>, I could put the norm of my plane as <-3/2, 2, - 1> since that would be parallel? Hence a = -3/4 and b = 1/4 ?
 
  • #5
riskybeats said:
The norm of my plane is influenced by a and b. This is what I have:

2a (X - a) + 8b (y - b) - z + a^2 + 4b^2 = 0

So if the vector is <3, -4, 2>, I could put the norm of my plane as <-3/2, 2, - 1> since that would be parallel? Hence a = -3/4 and b = 1/4 ?


Hm, could you post the equation of the normal?
 
  • #6
boboYO said:
Hm, could you post the equation of the normal?

Sorry but I'm not sure what you mean by the equation of the normal. Do you mean the vector equation of the tangent plane?

<-3/2, 2, -1>[tex]\cdot[/tex]<x + 3/4, y - 1/4, z> = 0?

Although I do still have those other numbers a2 + 4b2. Would that influence the normal of my plane still? I am tired and will probably think about this better later. Thanks for the help so far!
 

What is the definition of a plane perpendicular to a vector?

A plane perpendicular to a vector is a flat, two-dimensional surface that intersects the vector at a 90 degree angle, creating a right angle. This means that the vector is parallel to the plane's normal vector, which is a vector that is perpendicular to the plane and points outward from its surface.

How do you determine if a plane is perpendicular to a given vector?

To determine if a plane is perpendicular to a given vector, you can use the dot product. The dot product of a vector and the normal vector of the plane should equal 0 if the plane is perpendicular to the vector. Alternatively, you can also check if the angle between the vector and the normal vector is 90 degrees using the formula cos(theta) = (a * b) / (|a| * |b|), where a and b are the vectors and theta is the angle between them.

What is the equation of a plane perpendicular to a vector?

The equation of a plane perpendicular to a vector is ax + by + cz = d, where (a,b,c) is the normal vector of the plane, (x,y,z) are the coordinates of a point on the plane, and d is a constant. This equation can also be written in vector form as n * (r - r0) = 0, where n is the normal vector, r is the position vector of a point on the plane, and r0 is the position vector of a known point on the plane.

Can a plane be perpendicular to more than one vector?

Yes, a plane can be perpendicular to more than one vector. This means that the plane is parallel to all of the normal vectors of the vectors it is perpendicular to. In three-dimensional space, a plane can be perpendicular to an infinite number of vectors as long as they are all parallel to each other.

What is the significance of planes perpendicular to vectors in physics and engineering?

Planes perpendicular to vectors are important in physics and engineering because they represent the direction of forces and can be used to calculate the magnitude and direction of those forces. These planes are also used to determine the orientation of objects and their motion, which is crucial in fields such as mechanics, aerodynamics, and structural engineering.

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