Problem 32
Question
Find the distance from the point (-2,1,4) to the a. plane \(x=3\) b. plane \(y=-5\) c. plane \(z=-1\)
Step-by-Step Solution
Verified Answer
a. 5; b. 6; c. 5.
1Step 1: Understanding the distance to the x-plane
The problem asks us to find the distance from a point \((-2,1,4)\) to the plane \(x=3\). Planes parallel to the yz-plane have a constant x-coordinate. The distance from the point (-2,1,4) to the plane \(x=3\) is simply the absolute difference in the x-coordinates of the point and the plane, because they lie along a line parallel to the x-axis.
2Step 2: Calculating the x-distance
To find the distance from the point \((-2, 1, 4)\) to the plane \(x=3\), subtract the x-coordinate of the point from the x-value of the plane and take the absolute value. \(\text{Distance}= |3 - (-2)| = |3 + 2| = 5.\)
3Step 3: Understanding the distance to the y-plane
The plane \(y=-5\) is parallel to the xz-plane and has a constant y-coordinate. Similarly, the distance from the point \((-2,1,4)\) to the plane \(y=-5\) depends only on the difference in y-coordinates.
4Step 4: Calculating the y-distance
To find the distance from the point \((-2, 1, 4)\) to the plane \(y=-5\), subtract the y-coordinate of the point from the y-value of the plane and take the absolute value: \(\text{Distance}= |-5 - 1| = |-6| = 6.\)
5Step 5: Understanding the distance to the z-plane
The plane \(z=-1\) is parallel to the xy-plane and has a constant z-coordinate. The z-distance is the difference between the z-coordinate of the point and the z-coordinate of the plane.
6Step 6: Calculating the z-distance
To find the distance from the point \((-2, 1, 4)\) to the plane \(z=-1\), subtract the z-coordinate of the point from the z-value of the plane and take the absolute value: \(\text{Distance}= |-1 - 4| = |-5| = 5.\)
Key Concepts
Coordinate Geometry3D GeometryPlane Equations
Coordinate Geometry
Coordinate geometry, often referred to as Cartesian geometry, is a fundamental branch of mathematics where geometric problems are solved using a coordinate system. This system utilizes coordinates on planes to define points and lines. It plays a crucial role when we want to represent points like \((-2, 1, 4)\) in a three-dimensional space.
- Coordinates: A point in 3D space is defined by three numbers \((x, y, z)\), each representing a distance along the corresponding axis on the Cartesian plane.
- Distance Calculations: To find the distance between a point and a plane, you mainly identify differences in one coordinate, while the other two remain constant.
3D Geometry
In 3D geometry, we explore shapes and constructs in space defined by three dimensions: length, width, and height. Each point in this space is designated by three coordinates.
Three-dimensional geometry allows us to perform an array of operations, crucial for both theoretical mathematics and practical applications.
- Visual Representation: Visualizing points, lines, and planes becomes significant when dealing with three-dimensional space.
- Distances: Unlike 2D, where distances are limited to horizontal or vertical, in 3D, distances account for depth alongside traditional length and width.
Plane Equations
A plane in mathematics is a flat, two-dimensional surface extending infinitely in 3D space. To fully understand the problem involving calculating the distance from a point to a plane, one must grasp plane equations and properties. In our instance, planes are showcased with equations like \(x=3\), \(y=-5\), and \(z=-1\). Knowing these will help determine their orientation and behavior in space.
- Standard forms: When a plane is written as \(x = c\), it represents all points for any y and z, but with a fixed x-value.
- Parallelism: Planes are always parallel to the axis that isn't in their equation. For instance, the plane \(x=3\) is parallel to the yz-plane.
Other exercises in this chapter
Problem 32
Find a plane through the points \(P_{1}(1,2,3), P_{2}(3,2,1)\) and perpendicular to the plane \(4 x-y+2 z=7\)
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Find the vectors whose lengths and directions are given. Try to do the calculations without writing. $$\begin{array}{ll} \text { Length } & \text { Direction }
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If \(\mathbf{u} \times \mathbf{v}=\mathbf{u} \times \mathbf{w}\) and \(\mathbf{u} \neq \mathbf{0}\) then does \(\mathbf{v}=\mathbf{w} ?\) Give reasons for your
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Sketch the surfaces ASSORTED $$z=1+y^{2}-x^{2}$$
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