Problem 78
Question
Find the equation of the plane each of whose points is equidistant from \((-2,1,4)\) and \((6,1,-2)\).
Step-by-Step Solution
Verified Answer
The plane equation is: \(4x - 3z + 5 = 0\).
1Step 1: Understand the Problem
We need to find a plane such that for any point \((x, y, z)\) on the plane, its distance to \((-2, 1, 4)\) is equal to its distance to \((6, 1, -2)\).
2Step 2: Write the Distance Formulas
The distance from a point \((x, y, z)\) to \((-2, 1, 4)\) is given by:\[\sqrt{(x + 2)^2 + (y - 1)^2 + (z - 4)^2}\] The distance to \((6, 1, -2)\) is:\[\sqrt{(x - 6)^2 + (y - 1)^2 + (z + 2)^2}\]
3Step 3: Set the Distances Equal
Since the point is equidistant, set the distances equal:\[\sqrt{(x + 2)^2 + (y - 1)^2 + (z - 4)^2} = \sqrt{(x - 6)^2 + (y - 1)^2 + (z + 2)^2}\]
4Step 4: Square Both Sides
To eliminate the square roots, square both sides of the equation:\[(x + 2)^2 + (y - 1)^2 + (z - 4)^2 = (x - 6)^2 + (y - 1)^2 + (z + 2)^2\]
5Step 5: Simplify the Equation
Cancel \((y - 1)^2\) from both sides, expand the other terms, and simplify:\[x^2 + 4x + 4 + z^2 - 8z + 16 = x^2 - 12x + 36 + z^2 + 4z + 4\]Simplifying further, we get:\[16x - 12z + 20 = 0\]
6Step 6: Rearrange to Find the Plane Equation
Rearrange the equation:\[4x - 3z + 5 = 0\]This is the equation of the plane where every point \((x, y, z)\) is equidistant from \((-2, 1, 4)\) and \((6, 1, -2)\).
Key Concepts
Equidistant PointsDistance Formula3D GeometryCoordinate Geometry
Equidistant Points
Equidistant points are points that have the same distance from two or more different locations. In the context of plane equations, we are interested in finding a plane where every point on it is equidistant from two given points. Imagine a "bridge" that positions itself perfectly between two cities on a map.
This bridge represents the plane, ensuring that any traveler can start at either city and reach the plane in the same amount of time.
In mathematical terms, this means using the distance formula to find those points which satisfy this condition, creating a geometrical balance or equilibrium.
This bridge represents the plane, ensuring that any traveler can start at either city and reach the plane in the same amount of time.
In mathematical terms, this means using the distance formula to find those points which satisfy this condition, creating a geometrical balance or equilibrium.
Distance Formula
The distance formula is a crucial tool used to calculate the straight-line distance between two points in space. In three-dimensional geometry, the formula takes this form:
It helps us find critical elements of the geometry of shapes like cubes, spheres, and, crucially, the planes that slice through 3D space. By applying the distance formula between point \((x, y, z)\) on the plane and given points, we establish connections essential for crafting the plane's equation.
- The distance between points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\) is given by \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).
It helps us find critical elements of the geometry of shapes like cubes, spheres, and, crucially, the planes that slice through 3D space. By applying the distance formula between point \((x, y, z)\) on the plane and given points, we establish connections essential for crafting the plane's equation.
3D Geometry
3D Geometry involves understanding shapes and structures in three-dimensional space, which includes length, width, and height. This area of geometry is rich and multi-faceted, dealing with concepts beyond our usual 2-dimensional understanding.
In the realm of planes, which are flat surfaces stretching indefinitely in 3D space, equations describe their orientation and position.
The task in 3D geometry is often to connect this visual spatial understanding with algebraic representations, like plane equations.
Any point in 3D space is represented by a set of coordinates \( (x, y, z) \), mapping its unique position.
In the realm of planes, which are flat surfaces stretching indefinitely in 3D space, equations describe their orientation and position.
The task in 3D geometry is often to connect this visual spatial understanding with algebraic representations, like plane equations.
Any point in 3D space is represented by a set of coordinates \( (x, y, z) \), mapping its unique position.
Coordinate Geometry
Coordinate Geometry (or Analytic Geometry) combines algebra and geometry, allowing points, lines, and planes to be represented using equations. It's the language used to convert spatial dimensions into interpretable data using coordinates.
Consider the way a map can express locations using longitude and latitude - in coordinate geometry, every point in space is described similarly with sets of numbers.
The power lies in transforming visual shapes into algebraic formulas, which we can solve or manipulate mathematically.
By finding the plane's equation where all points are equidistant from two given points, coordinate geometry simplifies the complex labor of visualizing and calculating distances and positions in 3D space.
Consider the way a map can express locations using longitude and latitude - in coordinate geometry, every point in space is described similarly with sets of numbers.
The power lies in transforming visual shapes into algebraic formulas, which we can solve or manipulate mathematically.
By finding the plane's equation where all points are equidistant from two given points, coordinate geometry simplifies the complex labor of visualizing and calculating distances and positions in 3D space.
Other exercises in this chapter
Problem 77
Show that the curvature of the polar curve \(r^{2}=\cos 2 \theta\) is directly proportional to \(r\) for \(r>0\).
View solution Problem 77
Find the distance from the sphere \(x^{2}+y^{2}+z^{2}+2 x+\) \(6 y-8 z=0\) to the plane \(3 x+4 y+z=15\).
View solution Problem 79
Draw the graph of \(x=4 \cos t, y=3 \sin (t+0.5)\), \(0 \leq t \leq 2 \pi .\) Estimate its maximum and minimum curvature by looking at the graph (curvature is t
View solution Problem 79
Prove the Cauchy-Schwarz Inequality for two-dimensional vectors: $$ |\mathbf{u} \cdot \mathbf{v}| \leq\|\mathbf{u}\|\|\mathbf{v}\| $$
View solution