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Chapter 11: Problem 3

Find an equation of the plane that passes through the point \(P\) and has the vector \(\mathbf{n}\) as a normal. $$ P(2,6,1) ; \mathbf{n}=\langle 1,4,2\rangle $$

Short Answer

Expert verified
The equation of the plane is \(x + 4y + 2z = 28\).

Step by step solution

01

Understand the Normal Vector

The normal vector to the plane is given as \(\mathbf{n} = \langle 1, 4, 2 \rangle\). This means that the coefficients \(a, b,\) and \(c\) in the plane equation \(ax + by + cz = d\) are 1, 4, and 2, respectively.
02

Use the Given Point

We have a point \(P(2, 6, 1)\) through which the plane passes. This point satisfies the plane equation \(1 \times x + 4 \times y + 2 \times z = d\). Substituting \(x = 2\), \(y = 6\), and \(z = 1\) into the equation gives us:\[1(2) + 4(6) + 2(1) = d\]
03

Solve for d

Calculate the right-hand side of the equation:\[1(2) + 4(6) + 2(1) = 2 + 24 + 2 = 28\]Thus, \(d = 28\).
04

Write the Plane Equation

Substitute \(a = 1\), \(b = 4\), \(c = 2\), and \(d = 28\) back into the plane equation to get the final equation of the plane:\[x + 4y + 2z = 28\].

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Key Concepts

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

Understanding the Normal Vector
The concept of a normal vector is fundamental when discussing the equation of a plane. A normal vector is a vector that is perpendicular to the plane. It defines the plane's orientation in three-dimensional space. In this scenario, the normal vector is given as \(\mathbf{n} = \langle 1, 4, 2 \rangle\).

Essentially, this means that the coefficients of \(x\), \(y\), and \(z\) in the plane equation, \(ax + by + cz = d\), are the same as the components of the normal vector \(\mathbf{n}\).

For our problem, the components \(a\), \(b\), and \(c\) are \(1\), \(4\), and \(2\) respectively. These numbers dictate how the plane is oriented spatially.
The Role of a Point in the Plane
To write an equation for a plane, you need a point that lies on the plane. This point provides a specific location that the plane must pass through. Our given point is \(P(2, 6, 1)\).

This point ensures that the plane equation is accurate and not just a theoretical plane floating somewhere. For a point \((x, y, z)\) to lie on the plane, it must satisfy the equation determined by the normal vector.
  • With \(P(2, 6, 1)\), plug \(x = 2\), \(y = 6\), and \(z = 1\) into the plane equation \(1x + 4y + 2z = d\).
  • This step is crucial for finding the constant \(d\) which completes the equation of our specific plane.
Formulating the Equation of a Plane
Once you have the normal vector and a point through which the plane passes, formulating the equation of the plane becomes straightforward. The general form of a plane equation is \(ax + by + cz = d\).

In our case, the normal vector \(\mathbf{n} = \langle 1, 4, 2 \rangle\) provides the coefficients \(1\), \(4\), and \(2\).

Substitute the point \(P(2, 6, 1)\) into the equation, resulting in:
  • \(1(2) + 4(6) + 2(1) = d\)
  • This simplifies to \(2 + 24 + 2 = 28\)
Hence, \(d = 28\), making the final equation of the plane \(x + 4y + 2z = 28\). This complete equation defines a specific plane in three dimensional space and ensures it is both oriented correctly and passes through a known point.

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Most popular questions from this chapter

Sketch the surface. $$ z=\sqrt{x^{2}+y^{2}-1} $$

These exercises refer to the hyperbolic paraboloid \(z=y^{2}-x^{2}\) (a) Find an equation of the parabolic trace in the plane \(y=2\) (b) Find the vertex of the parabola in part (a). (c) Find the focus of the parabola in part (a). (d) Describe the orientation of the focal axis of the parabola in part (a) relative to the coordinate axes.

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