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Chapter 9: Problem 14

Two fixed points, \(A\) and \(B\), in three-dimensional space have position vectors \(\mathbf{a}\) and b. Identify the plane \(P\) given by $$ (\mathbf{a}-\mathbf{b}) \cdot \mathbf{r}=\frac{1}{2}\left(a^{2}-b^{2}\right) $$ where \(a\) and \(b\) are the magnitudes of a and \(\mathbf{b}\). Show also that the equation $$ (\mathbf{a}-\mathbf{r}) \cdot(\mathbf{b}-\mathbf{r})=0 $$ describes a sphere \(S\) of radius \(|\mathbf{a}-\mathbf{b}| / 2\). Deduce that the intersection of \(P\) and \(S\) is also the intersection of two spheres, centred on \(A\) and \(B\), and each of radius \(|\mathbf{a}-\mathbf{b}| / \sqrt{2}\)

Short Answer

Expert verified
The plane P has a normal vector of \(\mathbf{a}-\mathbf{b}\). The radius of sphere S is \(\frac{|\mathbf{a}-\mathbf{b}|}{2}\). The intersection of P and S are also the intersection of two spheres centred on A and B each with a radius of \(|\mathbf{a}-\mathbf{b}| / \sqrt{2}\).

Step by step solution

01

Identify the Plane

Given the equation associated with the plane: \[(\mathbf{a}-\mathbf{b}) \cdot \mathbf{r}=\frac{1}{2}\left(a^{2}-b^{2}\right)\]This equation is in the form of the standard equation of a plane \(\mathbf{n}\cdot\mathbf{r}=d\) where \(\mathbf{n}\) is the normal vector to the plane, \(\mathbf{r}\) is a generic position vector and \(d\) is the perpendicular distance from the origin to the plane. From this, it can be seen that the normal to the plane is \(\mathbf{a}-\mathbf{b}\).
02

Show Sphere Equation

Given the second equation: \[(\mathbf{a}-\mathbf{r}) \cdot(\mathbf{b}-\mathbf{r})=0\]This is the standard equation of a sphere. To prove that the radius is \(\frac{|\mathbf{a}-\mathbf{b}|}{2}\), it's crucial to rearrange the sphere equation by expanding the dot product and simplifying the equation to obtain the expression for the radius.
03

Find Intersection of Spheres

To deduce that the intersection of P and S is also the intersection of two spheres, centred on A and B, with radius \(|\mathbf{a}-\mathbf{b}| / \sqrt{2}\), it is key to take given plane equation and sphere equation into consideration and isolate \(\mathbf{r}\) in both equations. Once this is done, the two expressions for \(\mathbf{r}\) can equated and solved for radius to get the desired result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Calculus
Vector calculus is a branch of mathematics concerned with vector fields and their operations. It allows us to handle and manipulate vectors in multiple dimensions, which is essential for analyzing physical phenomena like forces and fields.
Vectors have both magnitude and direction. In our exercise, we use position vectors for points in space, represented by \( \mathbf{a} \) and \( \mathbf{b} \).
This forms the backbone of many geometric analyses because vectors make it possible to articulate directions and displacements concisely.
Key operations in vector calculus include:
  • Dot Product: Measures the degree to which two vectors are parallel. Mathematically, it's calculated as \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta \), where \( \theta \) is the angle between \( \mathbf{u} \) and \( \mathbf{v} \).
  • Cross Product: Produces a vector perpendicular to two given vectors. Useful for finding angles and generating normal vectors to planes.
  • Magnitude: Gives the length of a vector, calculated using \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).
Understanding these operations is key to solving more complex geometric problems in 3D space.
Equation of Plane
The equation of a plane allows us to define a flat, infinite surface in three-dimensional space using the dot product from vector calculus. In our exercise, the plane equation is presented as:\[(\mathbf{a}-\mathbf{b}) \cdot \mathbf{r}=\frac{1}{2}(a^{2}-b^{2})\]Here, \( \mathbf{a} - \mathbf{b} \) represents the normal vector, which is perpendicular to the plane.
The normal vector plays a crucial role because:
  • It determines the orientation of the plane.
  • Every point \( \mathbf{r} \) on the plane satisfies the equation when substituted.
  • The equation simplifies to the standard form \( \mathbf{n} \cdot \mathbf{r} = d \), linking geometry with algebra.
Simplifying and understanding this form helps us analyze intersections with other geometric figures, forming a pathway to expand our problem-solving toolkit.
Equation of Sphere
The equation of a sphere encapsulates a three-dimensional shape where all points are equidistant from a central point. In our exercise, the sphere is given by:\[(\mathbf{a}-\mathbf{r}) \cdot(\mathbf{b}-\mathbf{r})=0\]This equation demonstrates that the sphere's center lies at the midpoint of \( \mathbf{a} \) and \( \mathbf{b} \), and the radius is \(|\mathbf{a}-\mathbf{b}| / 2\).
Important aspects of the sphere's equation include:
  • A sphere has rotational symmetry about its center.
  • The parameters \( \mathbf{a} \) and \( \mathbf{b} \) help define the sphere's position and size in 3D space.
  • Rewriting the equation through expansion shows all points \( \mathbf{r} \) maintain a constant distance from the center, proving its spherical nature.
Comprehending this representation is vital as it allows us to visualize and calculate other properties such as intersections with planes.
Intersection of Geometrical Figures
Intersecting geometric figures involve finding common points between different shapes, such as planes and spheres. This is a cornerstone in analytical geometry as it permits modifying and understanding spatial relations.
The exercise asks us to find the intersection between a plane and a sphere and further deduce their relations. Here is the process:
  • Set the plane's equation equal to the sphere's equation, effectively solving for the common points \( \mathbf{r} \).
  • In this exercise, two additional spheres centered at points \( A \) and \( B \) of radius \( |\mathbf{a}-\mathbf{b}| / \sqrt{2} \) intersect.
  • These insights are logical steps built upon the properties of each shape and require meticulous algebraic handling.
This understanding offers a considerate perspective on synthesizing various geometric properties, resulting in more profound spatial insights.

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Most popular questions from this chapter

Prove Lagrange's identity, i.e. $$ (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}) $$

By proceeding as indicated below, prove the parallel axis theorem, which states that, for a body of mass \(M\), the moment of inertia \(I\) about any axis is related to the corresponding moment of inertia \(I_{0}\) about a parallel axis that passes through the centre of mass of the body by $$ I=I_{0}+M a_{\perp}^{2} $$ where \(a_{\perp}\) is the perpendicular distance between the two axes. Note that \(I_{0}\) can be written as $$ \int(\mathbf{n} \times \mathbf{r}) \cdot(\hat{\mathbf{n}} \times \mathbf{r}) d m $$ where \(r\) is the vector position, relative to the centre of mass, of the infinitesimal mass \(d m\) and \(\hat{\mathrm{n}}\) is a unit vector in the direction of the axis of rotation. Write a similar expression for \(I\) in which \(r\) is replaced by \(\mathbf{r}^{\prime}=\mathbf{r}-\mathbf{a}\), where \(\mathbf{a}\) is the vector position of any point on the axis to which \(I\) refers. Use Lagrange's identity and the fact that \(\int \mathbf{r} d m=0\) (by the definition of the centre of mass) to establish the result.

The vectors \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are coplanar and related by $$ \lambda \mathbf{a}+\mu \mathbf{b}+v \mathbf{c}=0 $$ where \(\lambda, \mu, v\) are not all zero. Show that the condition for the points with position vectors \(\alpha \mathbf{a}, \beta \mathbf{b}\) and \(\gamma \mathbf{c}\) to be collinear is $$ \frac{\lambda}{\alpha}+\frac{\mu}{\beta}+\frac{v}{\gamma}=0 $$

Let \(O, A, B\) and \(C\) be four points with position vectors \(\mathbf{0}, \mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\), and denote by \(\mathbf{g}=\lambda \mathbf{a}+\mu \mathbf{b}+v \mathbf{c}\) the position of the centre of the sphere on which they all lie. (a) Prove that \(\lambda, \mu\) and \(v\) simultaneously satisfy $$ (\mathbf{a} \cdot \mathbf{a}) \lambda+(\mathbf{a} \cdot \mathbf{b}) \mu+(\mathbf{a} \cdot \mathbf{c}) v=\frac{1}{2} a^{2} $$ and two other similar equations. (b) By making a change of origin, find the centre and radius of the sphere on which the points \(\mathbf{p}=3 \mathbf{i}+\mathbf{j}-2 \mathbf{k}, \mathbf{q}=4 \mathbf{i}+3 \mathbf{j}-3 \mathbf{k}, \mathbf{r}=7 \mathbf{i}-3 \mathbf{k}\) and \(\mathbf{s}=6 \mathbf{i}+\mathbf{j}-\mathbf{k}\) all lie.

Show that the points \((1,0,1),(1,1,0)\) and \((1,-3,4)\) lie on a straight line. Give the equation of the line in the form $$ \mathbf{r}=\mathbf{a}+\lambda \mathbf{b} $$
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