Chapter 9: Problem 25
Find the points on the given surface at which the tangent plane is parallel to the indicated plane. $$ x^{2}+y^{2}+z^{2}=7 ; 2 x+4 y+6 z=1 $$
Short Answer
Expert verified
The points are \((\frac{\sqrt{2}}{2}, \sqrt{2}, \frac{3\sqrt{2}}{2})\) and \((-\frac{\sqrt{2}}{2}, -\sqrt{2}, -\frac{3\sqrt{2}}{2})\).
Step by step solution
01
Understand the problem
To find the points on the surface \(x^2 + y^2 + z^2 = 7\) where the tangent plane is parallel to the plane \(2x + 4y + 6z = 1\), we need to find points where the normal vector to the surface is a scalar multiple of the normal vector of the given plane.
02
Find the normal vector to the given plane
The normal vector to the plane \(2x + 4y + 6z = 1\) is \(\langle 2, 4, 6 \rangle\).
03
Find the gradient of the surface
The gradient of the surface \(x^2 + y^2 + z^2 = 7\) is \(abla f = \langle 2x, 2y, 2z \rangle\), which represents the normal vector to the tangent plane at any point \((x, y, z)\) on the surface.
04
Set the gradient equal to a scalar multiple of the plane's normal vector
For the tangent plane to be parallel to \(2x + 4y + 6z = 1\), set \(abla f = k \cdot \langle 2, 4, 6 \rangle\), leading to \(\langle 2x, 2y, 2z \rangle = \langle 2k, 4k, 6k \rangle\).
05
Solve for the points on the surface
This gives the system of equations: 1. \(2x = 2k\) \(\Rightarrow x = k\)2. \(2y = 4k\) \(\Rightarrow y = 2k\)3. \(2z = 6k\) \(\Rightarrow z = 3k\)Substitute \(x = k\), \(y = 2k\), and \(z = 3k\) into \(x^2 + y^2 + z^2 = 7\):\[k^2 + (2k)^2 + (3k)^2 = 7\]\[k^2 + 4k^2 + 9k^2 = 7\]\[14k^2 = 7\]\[k^2 = \frac{7}{14} = \frac{1}{2}\]\[k = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}\]
06
Find the points using solved \(k\) values
If \(k = \frac{\sqrt{2}}{2}\), then \(x = \frac{\sqrt{2}}{2}\), \(y = \sqrt{2}\), \(z = \frac{3\sqrt{2}}{2}\).If \(k = -\frac{\sqrt{2}}{2}\), then \(x = -\frac{\sqrt{2}}{2}\), \(y = -\sqrt{2}\), \(z = -\frac{3\sqrt{2}}{2}\).
07
Confirm the points satisfy the surface equation
Check the points \((\frac{\sqrt{2}}{2}, \sqrt{2}, \frac{3\sqrt{2}}{2})\) and \((-\frac{\sqrt{2}}{2}, -\sqrt{2}, -\frac{3\sqrt{2}}{2})\) against the surface equation:1. For \((\frac{\sqrt{2}}{2}, \sqrt{2}, \frac{3\sqrt{2}}{2})\):\[\left(\frac{\sqrt{2}}{2}\right)^2 + (\sqrt{2})^2 + \left(\frac{3\sqrt{2}}{2}\right)^2 = \frac{2}{4} + 2 + \frac{18}{4} = \frac{1}{2} + 2 + \frac{9}{2} = 7\]2. For \((-\frac{\sqrt{2}}{2}, -\sqrt{2}, -\frac{3\sqrt{2}}{2})\), the same computation holds.Thus, both points are valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Surface Equation
A surface equation is a mathematical expression that defines a surface in three-dimensional space. In our exercise, the surface equation is given as \(x^2 + y^2 + z^2 = 7\).
This particular equation represents a sphere centered at the origin, with radius \(\sqrt{7}\). Each point \((x, y, z)\) that satisfies this equation lies on the surface of the sphere.
A useful way to think about surface equations is by their geometric shapes. These shapes provide a visual understanding of the areas we work with when determining properties like tangent planes or normal vectors.
This particular equation represents a sphere centered at the origin, with radius \(\sqrt{7}\). Each point \((x, y, z)\) that satisfies this equation lies on the surface of the sphere.
A useful way to think about surface equations is by their geometric shapes. These shapes provide a visual understanding of the areas we work with when determining properties like tangent planes or normal vectors.
- Surface equations often appear in forms like spheres, ellipsoids, cylinders, and others, each having its unique properties that describe its geometry.
- By understanding the type of geometric object an equation represents, one can infer many of its spatial properties.
Normal Vector
A normal vector to a surface or plane is a vector that is perpendicular to every tangent vector at a given point on the surface.
For the plane \(2x + 4y + 6z = 1\), the normal vector is \(\langle 2, 4, 6 \rangle\). This means any direction parallel to this vector is perpendicular to the surface of the plane.
In the context of surfaces described by a surface equation, normal vectors are crucial as they help to establish parallelism and perpendicularity. Here are some key points:
For the plane \(2x + 4y + 6z = 1\), the normal vector is \(\langle 2, 4, 6 \rangle\). This means any direction parallel to this vector is perpendicular to the surface of the plane.
In the context of surfaces described by a surface equation, normal vectors are crucial as they help to establish parallelism and perpendicularity. Here are some key points:
- The direction of a normal vector matters—in terms of both gradients and specific plane descriptions.
- Finding a normal vector allows us to evaluate the tangent plane's orientation concerning other surfaces or planes.
Gradient
Gradients play a central role when dealing with surfaces and their tangent planes.
The gradient of a function \(f(x, y, z)\) provides the direction of maximum increase of the function, and it is a vector consisting of partial derivatives.
In our exercise, the gradient of the surface \(x^2 + y^2 + z^2 = 7\) is found as \(abla f = \langle 2x, 2y, 2z \rangle\). This gradient vector essentially acts as the normal vector to the tangent plane at any point on the surface. Moreover:
The gradient of a function \(f(x, y, z)\) provides the direction of maximum increase of the function, and it is a vector consisting of partial derivatives.
In our exercise, the gradient of the surface \(x^2 + y^2 + z^2 = 7\) is found as \(abla f = \langle 2x, 2y, 2z \rangle\). This gradient vector essentially acts as the normal vector to the tangent plane at any point on the surface. Moreover:
- The gradient informs us how steeply the function increases and the direction of that increase.
- It also determines the positioning of the tangent plane relative to the surface described by the function.