/*! 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 25 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether the following pairs of planes are parallel, orthogonal, or neither. $$x+y+4 z=10 \text { and }-x-3 y+z=10$$

Short Answer

Expert verified
Answer: The planes are orthogonal.

Step by step solution

01

Identify the normal vectors of the planes

To identify the normal vectors of the planes, we look at the coefficients of x, y, and z in each equation. For the first plane, the normal vector is given by \[\mathbf{n}_1 = (1, 1, 4)\] and for the second plane, the normal vector is given by \[\mathbf{n}_2 = (-1, -3, 1)\].
02

Check if the normal vectors are scalar multiples

To check if the vectors are scalar multiples, we will divide their respective components and see if the results are equal. \[\frac{-1}{1} = -1, \frac{-3}{1} = -3, \text{ and } \frac{1}{4} = 0.25\]Since the results are not equal, the vectors are not scalar multiples of each other, and therefore the planes are not parallel.
03

Check if the normal vectors are perpendicular

To check if the vectors are perpendicular, we will calculate their dot product. If the dot product is 0, the vectors are perpendicular. The dot product of \[\mathbf{n}_1 \text{ and } \mathbf{n}_2 \text{ is: } (1)(-1) + (1)(-3) + (4)(1) = -1 - 3 + 4 = 0\]Since the dot product is 0, the normal vectors are perpendicular, and therefore the planes are orthogonal.
04

Conclusion

The given planes with equations \(x + y + 4z = 10\) and \(-x - 3y + z = 10\) are orthogonal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Vectors
When you talk about planes in three-dimensional geometry, it’s crucial to understand the role of **normal vectors**. A normal vector is simply a vector that is perpendicular to a plane. It provides important information about the plane's orientation within the space.

Every plane surface can be represented mathematically by an equation of the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are constants. These constants play a significant role because they determine the normal vector of the plane. Specifically, the normal vector is \((a, b, c)\).

Let's consider an example: the plane described by the equation \(x + y + 4z = 10\). The normal vector for this plane is \((1, 1, 4)\). Similarly, for the plane \(-x - 3y + z = 10\), the normal vector is \((-1, -3, 1)\).

Normal vectors help to compare planes by determining how they relate to each other. They are key in understanding concepts like parallelism and orthogonality between planes.
Dot Product
The **dot product** is a mathematical operation that takes two equal-length sequences of numbers (often vectors) and returns a single number. It’s a fundamental tool for vector calculations, especially useful in determining angles between two vectors.

To calculate the dot product of two vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), you multiply their corresponding components and then sum up these products:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

The dot product is extremely helpful in determining whether two vectors are perpendicular to each other. For example, when finding if planes are orthogonal, like in our problem, you calculate the dot product of their normal vectors.

If the dot product equals zero, the vectors are perpendicular, indicating the planes are orthogonal. In our case, the normal vectors \((1, 1, 4)\) and \((-1, -3, 1)\) were found to have a dot product of 0:

\[(1)(-1) + (1)(-3) + (4)(1) = -1 - 3 + 4 = 0\]

This result confirmed that the planes are orthogonal.
Vector Perpendicularity
Understanding **vector perpendicularity** is key to grasping the relationship between geometric entities such as planes. When two vectors are perpendicular, it implies that they meet at a right angle. In the context of planes, this concept helps in determining if two planes are orthogonal.

For vectors \(\mathbf{u}\) and \(\mathbf{v}\) to be perpendicular, their dot product must be zero:

\[ \mathbf{u} \cdot \mathbf{v} = 0 \]

In the problem we considered, we checked for perpendicularity between normal vectors \((1, 1, 4)\) and \((-1, -3, 1)\). By computing their dot product, we found it to be zero, thus confirming their perpendicular nature.

Perpendicular vectors in three-dimensional space indicate that they span two individual directions that form no overlap, reflecting a clear 90-degree directional relationship. This property is used to verify that the two planes in question meet at a right angle, or are orthogonal. Understanding this principle is essential in solving geometry-related problems where vector relationships dictate the properties of more complex shapes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Extending Exercise \(62,\) when three electrical resistors with resistance \(R_{1}>0, R_{2}>0,\) and \(R_{3}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega, R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega,\) and \(R_{3}\) increases from \(1.5 \Omega\) to \(1.55 \Omega.\)

The temperature of points on an elliptical plate \(x^{2}+y^{2}+x y \leq 1\) is given by \(T(x, y)=25\left(x^{2}+y^{2}\right) .\) Find the hottest and coldest temperatures on the edge of the elliptical plate.

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$

Show that the following two functions have two local maxima but no other extreme points (therefore, there is no saddle or basin between the mountains). a. \(f(x, y)=-\left(x^{2}-1\right)^{2}-\left(x^{2}-e^{y}\right)^{2}\) b. \(f(x, y)=4 x^{2} e^{y}-2 x^{4}-e^{4 y}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.