Problem 42
Question
In Exercises \(41-46,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$P_{1}(-1,1,5), \quad P_{2}(2,5,0)$$
Step-by-Step Solution
Verified Answer
The distance between points is \(5\sqrt{2}\).
1Step 1: Understand the Distance Formula
To find the distance between two points in 3D space, use the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] where \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) are the coordinates of the two points.
2Step 2: Identify Point Coordinates
Identify the coordinates of the given points: \(P_1 = (-1, 1, 5)\) and \(P_2 = (2, 5, 0)\). Here, \(x_1 = -1\), \(y_1 = 1\), \(z_1 = 5\), \(x_2 = 2\), \(y_2 = 5\), and \(z_2 = 0\).
3Step 3: Apply the Distance Formula
Substitute the identified coordinates into the distance formula: \[ \text{Distance} = \sqrt{(2 - (-1))^2 + (5 - 1)^2 + (0 - 5)^2} \] Simplify inside the square root to find the distance.
4Step 4: Simplify the Expression
Calculate each component separately: \((2 - (-1))^2 = 3^2 = 9\), \((5 - 1)^2 = 4^2 = 16\), and \((0 - 5)^2 = (-5)^2 = 25\). Sum these values: \(9 + 16 + 25 = 50\).
5Step 5: Compute the Distance
Take the square root of the sum to find the distance: \[ \text{Distance} = \sqrt{50} \] Simplify to get \[ \text{Distance} = 5\sqrt{2} \].
Key Concepts
Euclidean distanceAnalytic geometryCoordinate geometry
Euclidean distance
The Euclidean distance is a fundamental concept in geometry. It's used to measure the straight-line distance between two points in space. In a 3D context, we typically represent these points with three coordinates each:
- Each point has an x (horizontal), y (vertical), and z (depth) coordinate.
- The formula for Euclidean distance between two points \(P_1(x_1, y_1, z_1)\)and \(P_2(x_2, y_2, z_2)\) is \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Analytic geometry
Analytic geometry, often referred to as coordinate geometry, is the study of geometry using a coordinate system. It uniquely combines algebra and geometry by using coordinates to solve geometric problems easily. This is accomplished by representing geometric shapes like lines, planes, and curves with algebraic equations. By converting geometric problems into algebraic equations:
- We can apply algebraic methods to obtain solutions for geometrical problems.
- This is useful in finding distances, as it allows us not only to work in two dimensions but also to extend our understanding to three dimensions and beyond.
Coordinate geometry
Coordinate geometry is a branch of mathematics that employs a coordinate plane to describe and solve geometric problems. It's essentially a more practical and applied approach to geometry, where:
- We use a grid-like system (often an x, y, and sometimes z-axis) to define the location of points.
- The relationship between points is analyzed using algebraic concepts.
Other exercises in this chapter
Problem 42
In Exercises \(39-44,\) find the distance from the point to the plane. $$ (2,2,3), \quad 2 x+y+2 z=4 $$
View solution Problem 42
Linear combination Let \(\mathbf{u}=\mathbf{i}-2 \mathbf{j}, \mathbf{v}=2 \mathbf{i}+3 \mathbf{j},\) and \(\mathbf{w}=\) \(\mathbf{i}+\mathbf{j} .\) Write \(\ma
View solution Problem 42
Sketch the surfaces in Exercises \(13-44.\) ASSORTED $$y^{2}-x^{2}-z^{2}=1$$
View solution Problem 43
Find the areas of the triangles whose vertices are given in Exercises \(41-47 .\) $$ A(-5,3), \quad B(1,-2), \quad C(6,-2) $$
View solution