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Chapter 7: Problem 39

In Problems, find an equation of the plane that contains the given point and is perpendicular to the indicated vector. $$ (5,1,3) ; 2 \mathrm{i}-3 \mathrm{j}+4 \mathrm{k} $$

Short Answer

Expert verified
Equation of the plane: \(2x - 3y + 4z = 19\).

Step by step solution

01

Understand the Equation of a Plane

The equation of a plane in 3D space can be written as \( ax + by + cz = d \), where \((a, b, c)\) is a normal vector to the plane, and \((x, y, z)\) is any point on the plane. \(d\) is a constant determined by a specific point on the plane.
02

Identify the Perpendicular Vector

From the problem, the vector \(2\mathrm{i} - 3\mathrm{j} + 4\mathrm{k}\) is given as perpendicular to the plane. This vector, \((a, b, c) = (2, -3, 4)\), will be the coefficients in the plane equation.
03

Substitute the Point into the Plane Equation

The plane passes through the point \((5, 1, 3)\). Using the equation form \(ax + by + cz = d\), substitute \(a = 2\), \(b = -3\), \(c = 4\), and the point \((x, y, z) = (5, 1, 3)\).
04

Calculate the Plane Equation Constant \(d\)

Substitute into the plane equation to find \(d\):\[2(5) - 3(1) + 4(3) = d\]Calculate: \[10 - 3 + 12 = d\]Thus, \[d = 19.\]
05

Write the Final Equation of the Plane

The equation of the plane is \[2x - 3y + 4z = 19.\] This equation uses the normal vector coefficients and the calculated \(d\).

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Key Concepts

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

Normal Vector
In the context of a plane in 3D space, a normal vector is a critical component. It is a vector that is perpendicular to every point on the plane, essentially defining the orientation of the plane.
Consider a plane in space. It can be likened to a flat piece of paper floating. The normal vector is an arrow jutting out of this paper, ensuring it's always vertical to it.
In mathematical terms, if we have a normal vector represented as
  • (a, b, c),
this means it directly influences the plane's equation, written as
  • ax + by + cz = d.
This relationship shows how closely the normal vector is tied to the plane's equation, making it indispensable for characterizing the plane's overall orientation.
Plane in 3D Space
A plane in 3D space is an infinite flat surface. You can think of it like a giant, endless sheet of paper extending in all directions within a three-dimensional environment. Understanding a plane helps you comprehend how it interacts with other geometrical entities such as lines and points.
A plane is typically described using the equation:
  • ax + by + cz = d.
In this equation:
  • (a, b, c) are the coefficients from the normal vector, indicating the plane's tilt and rotation in space.
  • (x, y, z) represents any point lying on this plane.
  • d is a fixed value that shows where the plane sits relative to the origin.
So, when someone refers to a plane in 3D, they are talking about an infinite surface defined uniquely by these components, offering a foundational structure for 3D geometry.
Perpendicular Vector
A perpendicular vector in the context of a plane is particularly interesting and useful. Imagine a vector that sticks out perfectly at right angles to a plane, much like a pole standing perpendicular to a flat surface.
When determining the equation of a plane, this perpendicular vector becomes the normal vector. Its components are drawn directly into the equation of the plane as coefficients.
  • For example, if the given perpendicular vector is
    • (2, -3, 4),
    it will shape the plane's equation as
    • 2x - 3y + 4z = d.
Because it's perpendicular to the plane, this vector helps determine how the plane slices through the 3D space, thereby playing a crucial role in defining the spatial orientation of the plane.

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Most popular questions from this chapter

In Problems, find an equation of the plane that satisfies the given conditions. Contains \((2,4,8)\) and is perpendicular to the line \(x=10-3 t\), \(y=5+t, z=6-\frac{1}{2} t\)

Use the dot product to prove the triangle inequality \(\|\mathbf{a}+\mathbf{b}\| \leq\|\mathbf{a}\|+\|\mathbf{b}\| .\) [Hint: Consider property \((v i)\) of the dot product.]

In Problems, find, if possible, an equation of a plane that contains the given points. $$ (0,0,0),(1,1,1),(3,2,-1) $$

Find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\langle 5,7,2\rangle $$

The vectors \(p_{1}(x)=x+1, p_{2}(x)=x-1\) form a basis for the vector space \(P_{1}\) (a) Show that \(p_{1}(x)\) and \(p_{2}(x)\) are linearly independent. (b) Express the vector \(p(x)=5 x+2\) as a linear combination of \(p_{1}(x)\) and \(p_{2}(x)\).
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