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I assume you mean the normal vector in the plane of the circle
If you write the circle in the form f(x,y,z) = 0 e.g. x^2 + y^2 - r^2 = 0
then grad(f) gives you the normal vector (outward pointing). In cartesian (x,y,z) coordinates:
grad(f) = (df/dx, df/dy, df/dz)
So in our example:
grad(f) = (2x, 2y, 0)
This is the normal vector and is necessarily in the plane of the circle, even if this method is followed for a circle with some angle to the x-y plane :)
This works for any function of the form f(...) = 0, not just circles...
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Is surface Area a scalar or vector quantity?
This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point.Now imagine standing on the sheet of paper, the direction your head points to will always be a way to know how this plane is oriented in space. This is called the "normal" vector to this plane, it is at a right angle to the plane.If you now choos the convention to have the length of this vector ("the norm") equal to the area of this surface, you get a complete description of the two dimensional plane, its orientation in three dimensional space (the vector part) and how big this plane is (the length of this vector).Mathematically, you can express this by the "cross product"c ⃗=a⃗×b⃗whose magnitude is defined as|c|=|a||b|sinθwhich is equal to the area of the parallelogram those to vectors (which really define a plane) span. To steal this picture from wikipedia's article on the cross product:As I said in the beginning this is a very special thing for three dimensions, in higher dimensions, it doesn't work as neatly for various reasons. If you want to learn more about this topic a keyword would be "exterior algebra"Update:As for the physical significance of this concept, prominent examples are vector fields flowing through surfaces. Take a circular wire. This circle can be oriented in various ways in 3D. If you have an external magnetic field, you might know that this can induce an electric current, proportional to the rate of change of the amount flowing through the circle (think of this as how much the arrows perforate the area). If the magnetic field vectors are parallel to the circle (and thus orthogonal to its normal vector) they do not "perforate" the area at all, so the flow through this area is zero. On the other hand, if the field vectors are orthogonal to the plane (i.e. parallel to the normal), the maximally "perforate" this area and the flow is maximal.if you change the orientation of between those two states you can get electrical current.
What plane figure is not a polygon?
A circle?
What is the intersection of the sphere with the yz plane?
The intersection of a sphere with a plane is a point, or a circle.
What four shapes have formulas which include the symbol pi?
A circle,An ellipse, A sphere,A normal (Gaussian) distribution.A circle,An ellipse, A sphere,A normal (Gaussian) distribution.A circle,An ellipse, A sphere,A normal (Gaussian) distribution.A circle,An ellipse, A sphere,A normal (Gaussian) distribution.
What is all the points in a plane equidistant from the center?
That's a circle around the center, in the plane.
What is a vector which is orthogonal to the other vectors and is coplanar with the other vectors called?
In a plane, each vector has only one orthogonal vector (well, two, if you count the negative of one of them). Are you sure you don't mean the normal vector which is orthogonal but outside the plane (in fact, orthogonal to the plane itself)?
How to find orthogonal vector?
Given one vector a, any vector that satisfies a.b=0 is orthogonal to it. That is a set of vectors defining a plane orthogonal to the original vector.The set of vectors defines a plane to which the original vector a is the 'normal'.
How many coordinates are required to define a hexagonal vector or plane?
You only need 2 if you translate a normal 2D vector into a hexagonal one, using a formula that sets 2 of the hexagonal directions to mimic a normal coordinate plane. I learned by typing Hexagonal Chess into a search engine.
How do you use osculation in a sentence?
Depends on which osculation you mean: kissing or mathematics The word is a noun in both situations. You can have the moment or place of osculation, or in mathematics the plane or circle of osculation. "The phone rang just before osculation could occur." "Before he could find the plane of osculation, he needed to get the normal vector."
Is a circle a plane shape?
A circle is not a plane shape. A circle is round a plane is not. A plane has parts that are similar to a circle shape.
How do you find a normal vector?
A normal vector is a vector that is perpendicular or orthogonal to another vector. That means the angle between them is 90 degrees which also means their dot product if zero. I will denote (a,b) to mean the vector from (0,0) to (a,b) So let' look at the case of a vector in R2 first. To make it general, call the vector, V=(a,b) and to find a vector perpendicular to v, i.e a normal vector, which we call (c,d) we need ac+bd=0 So say (a,b)=(1,0), then (c,d) could equal (0,1) since their dot product is 0 Now say (a,b)=(1,1) we need c=-d so there are an infinite number of vectors that work, say (2,-2) In fact when we had (1,0) we could have pick the vector (0,100) and it is also normal So there is always an infinite number of vectors normal to any other vector. We use the term normal because the vector is perpendicular to a surface. so now we could find a vector in Rn normal to any other. There is another way to do this using the cross product. Given two vectors in a plane, their cross product is a vector normal to that plane. Which one to use? Depends on the context and sometimes both can be used!
Cross product is not difine in two space why?
When performing the cross product of two vectors (vector A and vector B), one of the properites of the resultant vector C is that it is perpendicular to both vectors A & B. In two dimensional space, this is not possible, because the resultant vector will be perpendicular to the plane that A & B reside in. Using the (i,j,k) unit vector notation, you could add a 0*k to each vector when doing the cross product, and the resultant vector will have zeros for the i & jcomponents, and only have k components.Two vectors define a plane, and their cross product is always a vector along the normal to that plane, so the three vectors cannot lie in a 2D space which is a plane.
What is lying in the same plane?
lying on the same plane indicates that the normal vector to that plane will always be the same i.e to that point or vector. Another way would be that it should satisfy the standard equation of plane Ax+By+Cz = D in the case of a point or the dot product should equal zero. <A,B,C> . <x-x0, y-y0, z-z0> = 0
What are the tangential and normal components of acceleration?
The tangential component of acceleration is the change in speed along the direction of motion, while the normal component is the change in direction of velocity. In other words, tangential acceleration affects the speed of an object, while normal acceleration affects the direction of motion.
How do you find the normal vector of a ball or sphere?
The normal vector to the surface is a radius at the point of interest.
What are the Examples of a vector quantity?
Some examples of a vector quantity would be a car or a plane.
Ten vectors addtogether to give zero resultant it is possible that nine of this vectors are in the same plane but the tenth not on this plane?
No. Let's assume the plane has coordinates x and y; the vector outside the plane has a component for the z-coordinate. In that case, another vector (or several) must also have a component in the z-coordinate, to compensate.No. Let's assume the plane has coordinates x and y; the vector outside the plane has a component for the z-coordinate. In that case, another vector (or several) must also have a component in the z-coordinate, to compensate.No. Let's assume the plane has coordinates x and y; the vector outside the plane has a component for the z-coordinate. In that case, another vector (or several) must also have a component in the z-coordinate, to compensate.No. Let's assume the plane has coordinates x and y; the vector outside the plane has a component for the z-coordinate. In that case, another vector (or several) must also have a component in the z-coordinate, to compensate.
