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Chapter 13: Problem 27

Find an equation for the plane through the origin containing the points (1,3,0) and (2,4,1)

Short Answer

Expert verified
The equation of the plane is \(3x - y - 2z = 0\).

Step by step solution

01

Identify the Normal Vector

To find the equation of a plane, we need a normal vector. First, identify two vectors lying on the plane using the given points. The vectors can be defined as \( \mathbf{a} = (1,3,0) \) and \( \mathbf{b} = (2,4,1) \). The normal vector \( \mathbf{n} \) is obtained by the cross product of \( \mathbf{a} \) and \( \mathbf{b} \).
02

Compute the Cross Product

The cross product \( \mathbf{n} = \mathbf{a} \times \mathbf{b} \) is calculated as follows: \[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 3 & 0 \2 & 4 & 1\end{vmatrix} = \mathbf{i}(3 \cdot 1 - 0 \cdot 4) - \mathbf{j}(1 \cdot 1 - 0 \cdot 2) + \mathbf{k}(1 \cdot 4 - 3 \cdot 2)\]\[= \mathbf{i} (3) - \mathbf{j} (1) + \mathbf{k} (-2) = (3, -1, -2)\]
03

Write the Equation of the Plane

The general equation of a plane is given by \( ax + by + cz = d \). Given that our plane passes through the origin \((0,0,0)\), \(d = 0\). Substitute \((a, b, c) = (3, -1, -2)\):\[ 3x - y - 2z = 0 \]

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Key Concepts

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

Cross Product
In three-dimensional space, the cross product is a useful operation to find a vector that is perpendicular to two given vectors. This is particularly useful when determining the normal vector for a plane, as in the above exercise.
  • The cross product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is denoted as \( \mathbf{a} \times \mathbf{b} \).
  • It results in a new vector \( \mathbf{n} = (n_1, n_2, n_3) \), where each component is derived from the determinant of a matrix involving \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the unit vectors.
  • The formula is:\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3\end{vmatrix} \]
  • Each component \( n_1, n_2, n_3 \) is calculated as a minor determinant of the matrix, influenced by switching certain elements.
For example, to find the normal vector to a plane that includes vectors \( \mathbf{a} = (1, 3, 0) \) and \( \mathbf{b} = (2, 4, 1) \), calculate the cross product as shown in the solution above.
Normal Vector
A normal vector to a plane is essential for defining the orientation or "tilt" of that plane in three-dimensional space. It is a vector perpendicular to every vector lying in the plane.
  • Once you have a normal vector \( \mathbf{n} \), you can uniquely determine the equation of the plane.
  • In our context, the normal vector comes from the cross product of two non-parallel vectors lying on the plane.
  • It can be represented as \( \mathbf{n} = (a, b, c) \), with these three coefficients defining the plane equation \( ax + by + cz = d \).
For our problem, where the vectors \( \mathbf{a} = (1,3,0) \) and \( \mathbf{b} = (2,4,1) \) lie on the plane, the calculated normal vector is \( \mathbf{n} = (3,-1,-2) \). This vector helps formulate the equation of the plane, assuming it passes through a given point, here the origin, simplifying to \( 3x - y - 2z = 0 \).
Vectors in 3D
Vectors in 3D space are essential mathematical tools that allow for the representation and manipulation of objects in three-dimensional physics and geometry.
  • A vector in 3D is expressed in the form \( \mathbf{v} = (x, y, z) \), with each component indicating the vector's magnitude along the x, y, and z axes.
  • Vectors can be visualized as arrows where direction and magnitude are clearly represented, enabling one to consider geometric properties and relationships.
  • Both addition and subtraction of vectors are straightforward, treating isotonic components (x with x, y with y, etc.).
  • When dealing with planes, any two non-parallel vectors lying on the plane can be used to calculate a normal vector, to define that plane.
In this exercise, two points (1,3,0) and (2,4,1) provide vectors \( \mathbf{a} \) and \( \mathbf{b} \), used to explore the plane's characteristics through the cross product and the resultant normal vector.

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Most popular questions from this chapter

For a fixed two-dimensional vector \(\vec{a}\) with \(\|\vec{a}\|=2\) determine how many vectors there are with \(\|\vec{b}\|=4\) and \(\vec{a} \cdot \vec{b}=4\)

Find a normal vector to the plane. $$2(x-z)=3(x+y)$$

Given \(\vec{v}=3 \vec{i}+4 \vec{j}\) and force vector \(\vec{F}\) find: (a) The component of \(\vec{F}\) parallel to \(\vec{v}\). (b) The component of \(\vec{F}\) perpendicular to \(\vec{v}\). (c) The work, \(W\), done by force \(\vec{F}\) through displacement \(\vec{v}\). $$\vec{F}=-3 \vec{i}-5 \vec{j}$$

Give reasons for your answer. The triangle in 3 -space with vertices (1,1,0),(0,1,0) and (0,1,1) has a right angle.

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