Determine an equation, the coordinates of the center, and the radius, of the sphere through the

Determine an equation, the coordinates of the center, and...

Key Concepts

Solving Systems of Equations in Three Dimensions

When multiple points on a sphere are provided, their substitution into the sphere's equation yields a system of algebraic equations. Solving this system, often using methods like elimination or matrix techniques, allows one to determine the sphere's parameters. This approach underpins many problems in analytic geometry where the objective is to determine unknown constants from geometric constraints.

Determining the Center and Radius

To find the center and radius of a sphere given several points on its surface, one substitutes the coordinates of these points into the sphere’s equation. This procedure results in a system of equations that can be solved simultaneously to find the unknown center coordinates (h, k, l) and the value of r². Understanding this process is key in tasks involving the analysis and synthesis of spherical equations.

Standard Equation of a Sphere

The standard form of a sphere's equation in three-dimensional space is expressed as (x-h)² + (y-k)² + (z-l)² = r², where (h, k, l) represents the coordinates of the sphere's center and r is its radius. This equation is fundamental in analytic geometry and provides a concise way to describe all points lying on the surface of a sphere.

Transcript

00:01 Hi there, so for this problem we are told let all a b and c be four points with position vectors so this position vectors will be 0 a b and c okay, and denotate by g the vector g is equal to lambda some constant times the vector a plus another constant mu times the vector b plus another constant new with the vector c the position of the center of the sphere on which they all lie.

00:43 So for part a of this problem we need to prove that mu lambda mu and nu simultaneously satisfy the expression that we are given and to order similar equations okay, so we know that each of the points and o a b and c is the same distance from the center g of the sphere in particular we know that the distance between o and g is equal to the distance between o and a so we will have then that the difference between 0 and the vector g and the vector 0 is equal to a minus the vector g and of course both sides to the square and this will be g to the square is equal to a to the square minus 2 times the dot product between these two vectors this plus g to the square the magnitude of the vector g to the square now in here we can solve we can see that this cancels this so solving in here for this dot product we will obtain that this is just 1 divided by 2 times a s square so with that said we can substitute what is the definition that we are given for g? so we will have a dot product with a plus mu times b plus b times c this is equal to 1 divided by 2 times a s square now we just do the dot product of this with each term in here so we will have the dot product of a with a this times lambda plus the dot product of a with b this times mu this plus the dot product of a times with c this times nu and this is equal to 1 divided by 2 times a s square so as you can see we have obtained this expression and you will obtain two similar equations that can be obtained from zero og which is equal to the distance between ob and og with oc you will just do the same procedure and will obtain something similar now for part b of this problem by making a change of a region we need to find the center and radius of the sphere on which the point so we are given some points so that will be 3 in the x direction for the vector p 1 in the y direction minus 2 in the z direction and the vector q is 4 in the x direction 3 in the y direction and 3 in the z direction the vector r that is equal to 7 in the x direction 3 in the z direction and the vector s is just 6 in the x direction 1 in the y direction and minus 1 in the z direction okay, so to use the previous result we make p say the origin of a new coordinate system in which we will have that this vector let's label this as p prime is just the vector p minus p.

04:12 So this will give us the vector just simply 0 0 0 because it's just the same vector now the vector q prime now we do the difference between the vector q and the vector p so when we do the difference between these two values you could see that the the vector that we obtain is 1 2 and minus 1 now we do the vector prime r that is equal to the vector r minus p this will give us 4 so remember that we're doing the difference between this and this that will give us for the first term is 4 for the next term is minus 1 for the next term is minus 1 and finally the vector s prime so that will be as but your s minus the vector p this will give us 3 0 and 1 now the center the century a g of this sphere on which p q r and s lie is then given by the following expression...

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