/*! 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 52 The plane passes through the poi... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 52

The plane passes through the points \((4,2,1)\) and \((-3,5,7)\) and is parallel to the \(z\) -axis.

Short Answer

Expert verified
The equation of the plane is \( y = 2 \).

Step by step solution

01

Determine the Direction Vector

First, calculate the direction vector of the two given points, which is formed by the differences in x, y and z coordinates. So the direction vector \( \vec{AB} \) joining the points A(4,2,1) and B(-3,5,7) is defined as B - A = (-3-4, 5-2, 7-1) = (-7, 3, 6).
02

Identify the Normal Vector

Since the plane is stated to be parallel to the z-axis, its normal vector should be either parallel to the x-axis or the y- axis. Choose the y-axis for simplicity implying normal vector is (0,1,0).
03

Apply the Plane Equation

Using the point-normal form of the plane equation, which is \( (r - a) \cdot n = 0 \), where \( r \) is a point on the plane, \( a \) is a known point on the plane and \( n \) is the normal vector to the plane, and plugging in the known values, we find the equation of the plane to be \( y = 2 \).

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Key Concepts

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

Direction Vector
A direction vector is essential in understanding how a line or a segment is oriented in space. It gives us the direction from one point to another. When we talk about two points, such as A(4,2,1) and B(-3,5,7), we find the direction vector, \( \vec{AB} \), by subtracting the coordinates of point A from point B.
  • For the x-component, subtract: \(-3 - 4 = -7\)
  • For the y-component, subtract: \(5 - 2 = 3\)
  • For the z-component, subtract: \(7 - 1 = 6\)
Thus, the direction vector \( \vec{AB} \) is \((-7, 3, 6)\).
This vector shows us the direction from point A to point B, across the x, y, and z axes. Knowing this helps in many geometric and algebraic operations, including computing the equation of planes, lines, and other geometric constructs.
Normal Vector
The normal vector is a vector that's perpendicular to a surface or a plane. It's significant because it helps define the orientation of the plane. In this specific problem, our plane is parallel to the z-axis, meaning its normal vector is within the x-y plane.
To simplify, we opted for a normal vector parallel to the y-axis: \( (0,1,0) \). This choice implies no inclination along the x or z axes, simplifying computations.
  • A normal vector perpendicular to the plane will have no influence from the z component in this scenario.
  • This normal vector dictates that any change in the y direction will affect the plane equation.
Choosing a specific normal vector is crucial in determining the plane's equation. It is like choosing the engine of a car鈥攊t affects how the plane or line will behave in 3D space.
Plane Equation
The plane equation forms the fundamental description of a plane in 3D geometry. Here, we use the point-normal form of the plane equation: \((r - a) \cdot n = 0\).
In this formula:
  • \(r\) is the position vector of any arbitrary point on the plane, expressed as \((x, y, z)\).
  • \(a\) is the vector of a specific known point on the plane, such as (4,2,1).
  • \(n\) is the normal vector, which we determined as \((0,1,0)\).
To derive the plane's equation:- Use point \(A (4, 2, 1)\) and normal vector \(n (0, 1, 0)\).- Plug these vectors into the point-normal form: \((r - a) \cdot n = 0\), yielding the simplified equation \( y = 2 \).
This equation, \(y = 2\), represents a plane parallel to the z-axis. It indicates that for every point on this plane, the y-value remains constant at 2, showcasing the role of the normal vector in defining the plane's orientation in space.

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Most popular questions from this chapter

In Exercises 97 and 98 , sketch the vector \(v\) and write its component form. \(\mathrm{v}\) lies in the \(y z\) -plane, has magnitude 2 , and makes an angle of \(30^{\circ}\) with the positive \(y\) -axis.

The spherical coordinates of a point \((x, y, z)\) are unique.Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

Find the distance between the point and the line given by the set of parametric equations.\((1,-2,4) ; \quad x=2 t, \quad y=t-3, \quad z=2 t+2\)

Consider the vectors \(\mathbf{u}=\langle\cos \alpha, \sin \alpha, 0\rangle\) and \(\mathbf{v}=\langle\cos \beta, \sin \beta, 0\rangle\) where \(\alpha>\beta\). Find the dot product of the vectors and use the result to prove the identity \(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)

Find the point(s) of intersection (if any) of the plane and the line. Also determine whether the line lies in the plane.\(5 x+3 y=17, \quad \frac{x-4}{2}=\frac{y+1}{-3}=\frac{z+2}{5}\)
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