Let O, A, B and C be four points with position vectors 0, 𝐚, 𝐛 and 𝐜, and denote by 𝐠=λ𝐚+μ𝐛+v 𝐜

Let O, A, B and C be four points with position vectors 0, 𝐚,...

Key Concepts

Position Vectors

Position vectors are used to represent points in space relative to an origin. They encapsulate the coordinates of points in vector form, making it easier to work with them algebraically. In the study of geometry and spatial relationships, position vectors allow the expression of points as linear combinations of basis vectors, which is crucial in establishing relationships among points such as distances and angles.

Sphere Equation via Distance Equality

A sphere is defined as the set of all points that are equidistant from a given fixed point called the center. The condition of equal distance leads to equations that equate the squared distance between an arbitrary point on the sphere and the center to the square of the radius. In problems involving multiple points on a sphere, these relationships create a system of equations useful for determining unknown parameters like the sphere’s center and radius.

Dot Product in Geometric Context

The dot product is a fundamental operation in vector geometry that gives a scalar result and measures the magnitude of projection of one vector onto another. In the context of sphere equations, the dot product is used to convert distance squared formulas into manageable algebraic forms. Specifically, the representation of squared lengths as dot products (e.g., a · a) simplifies the manipulation of geometry-based equations that arise when setting up conditions for points to lie on a sphere.

Change of Origin

Changing the origin, or coordinate transformation, is a technique used to simplify geometric problems by translating the coordinate system. By choosing a more convenient origin, the equations governing the positions of points can be simplified, making it easier to determine key parameters such as the center and radius of a sphere. This method is particularly useful when dealing with translated or shifted objects in space, as it often reduces complex equations to simpler ones.

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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