/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 In Problems \(11-14\), find the ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems \(11-14\), find the equation of the plane through the given points. \((7,0,0),(0,3,0)\), and \((0,0,5)\)

Short Answer

Expert verified
The equation of the plane is: \(15x + 35y + 21z = 105\).

Step by step solution

01

Identify Vectors

First, we need to find two vectors on the plane. Using the given points \({(7,0,0)}\), \({(0,3,0)}\), and \({(0,0,5)}\), we can find:- Vector \( extbf{AB}\) from \( extbf{A} = (7,0,0)\) to \( extbf{B} = (0,3,0)\): \[ \textbf{AB} = \textbf{B} - \textbf{A} = (0, 3, 0) - (7, 0, 0) = (-7, 3, 0) \]- Vector \( extbf{AC}\) from \( extbf{A} = (7, 0, 0)\) to \( extbf{C} = (0, 0, 5)\): \[ \textbf{AC} = \textbf{C} - \textbf{A} = (0, 0, 5) - (7, 0, 0) = (-7, 0, 5) \]
02

Find the Normal Vector

To find the normal vector to the plane, we calculate the cross product of the vectors \( extbf{AB}\) and \( extbf{AC}\). Let \[ \textbf{n} = \textbf{AB} \times \textbf{AC} \]The cross product is calculated as follows:\[ \textbf{n} = \left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k} \ -7 & 3 & 0 \ -7 & 0 & 5 \end{array} \right| = (3 \times 5 - 0 \times 0)\textbf{i} - (-7 \times 5 - 0 \times -7)\textbf{j} + (-7 \times 0 - 3 \times -7)\textbf{k} \]This simplifies to:\[ \textbf{n} = (15)\textbf{i} - (-35)\textbf{j} + (21)\textbf{k} = (15, 35, 21) \]
03

Formulate the Plane Equation

The general equation of a plane with a normal vector \[ (a, b, c) \] and passing through a point \((x_0, y_0, z_0)\) is:\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]Substitute \((a, b, c) = (15, 35, 21)\) and \((x_0, y_0, z_0) = (7, 0, 0)\):\[ 15(x - 7) + 35(y - 0) + 21(z - 0) = 0 \]
04

Simplify the Plane Equation

Expanding and simplifying the equation:\[ 15x - 105 + 35y + 21z = 0 \]Bringing all terms to one side results in:\[ 15x + 35y + 21z = 105 \]This is the equation of the plane.

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Key Concepts

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

Vectors
Vectors are mathematical objects that have both a magnitude and a direction. These objects are fundamental in physics and engineering because they allow us to describe the position and movement of points in space. A vector is typically represented by an arrow, where the length represents its magnitude, and the arrowhead indicates its direction. For example, the vector from point \(A = (7, 0, 0)\) to \(B = (0, 3, 0)\) is derived by subtracting the coordinates of \(A\) from \(B\). This gives us \(\textbf{AB} = (0 - 7, 3 - 0, 0 - 0) = (-7, 3, 0)\). Another vector can be calculated similarly, for example, \(\textbf{AC} = (0, 0, 5) - (7, 0, 0) = (-7, 0, 5)\).
  • Vectors allow us to navigate and manipulate points in space.
  • They are essential in deriving equations, such as the equation of a plane.
  • Finding vectors between points is the first step in solving problems involving planes.
Cross Product
The cross product is a vector operation that is crucial in three-dimensional vector analysis. It takes two non-parallel vectors and produces a third vector that is perpendicular to both. This resultant vector is called the normal vector. To compute the cross product, we use the determinant of a matrix, constructed from the components of the vectors involved.
For the vectors \(\textbf{AB} = (-7, 3, 0)\) and \(\textbf{AC} = (-7, 0, 5)\), the cross product \(\textbf{AB} \times \textbf{AC}\) can be expressed as:
\[\textbf{n} = \left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k} \-7 & 3 & 0 \-7 & 0 & 5 \end{array} \right|\]
This results in \(\textbf{n} = (15, 35, 21)\).
  • The cross product is key to finding a vector that is normal to the plane.
  • This operation involves determinant calculation, often used in physics.
  • Understanding the cross product helps in visualizing vector interactions in space.
Normal Vector
A normal vector is a vector that is perpendicular to a surface. In the context of a plane, this vector is crucial because it defines the plane's orientation in three-dimensional space. The normal vector provides a way to establish the equation of a plane through its components.
The normal vector, obtained from the cross product of two vectors on the plane, \(\textbf{n} = (15, 35, 21)\), guides us in writing the plane's equation. Its components \((a, b, c)\) are directly used in the plane equation formula.
By firmly understanding the concept of a normal vector, you can solve complex problems involving coordinates and surfaces.
  • The normal vector simplifies the process of defining a plane.
  • It directly impacts the coefficients in the plane's equation.
  • Understanding its role is fundamental in coordinate geometry.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a branch of geometry that uses a coordinate system to explore and solve geometric problems. By placing geometric shapes and points within a coordinate system, it becomes easier to analyze their properties and relationships using algebraic methods.
In solving for the equation of a plane, coordinate geometry allows us to relate algebraic expressions to spatial dimensions. Using the general formula, \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), we substitute the components of the normal vector and a point on the plane to construct the specific equation. In this case:
\[15(x - 7) + 35(y) + 21(z) = 0\]
This simplifies to \(15x + 35y + 21z = 105\).
  • Coordinate geometry provides a bridge between geometry and algebra.
  • It is essential for converting geometric problems into solvable equations.
  • It aids in the clear representation of geometric principles in mathematical terms.

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