/*! 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 15 Find an equation of the plane th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find an equation of the plane that passes through the point \(P_{0}\) with a normal vector \(\mathbf{n}\). Find an equation of the plane that is parallel to the vectors \langle 1,0,1\rangle and \(\langle 0,2,1\rangle,\) passing through the point (1,2,3)

Short Answer

Expert verified
Answer: The equations of the planes are: 1. For the plane passing through point \(P_{0}\) with normal vector \(\mathbf{n}\): \((n_x)(x-x_{0})+(n_y)(y-y_{0})+(n_z)(z-z_{0})=0\) 2. For the plane parallel to the vectors \(\langle 1,0,1 \rangle\) and \(\langle 0,2,1 \rangle\) and passing through the point (1,2,3): \(-2x+y-2z+5=0\)

Step by step solution

01

Equation of plane with point and normal vector

To find the equation of a plane that passes through a point \(P_{0}=(x_{0},y_{0},z_{0})\) with a normal vector \(\mathbf{n}= \langle n_{x}, n_{y}, n_{z} \rangle\), we can use the formula: \((n_x)(x-x_{0})+(n_y)(y-y_{0})+(n_z)(z-z_{0})=0\). Plug the point \(P_{0}\) and the normal vector \(\mathbf{n}\) into the formula to find the equation of the plane.
02

Find the normal vector using the cross product of the given parallel vectors

To find the normal vector of the plane parallel to the given vectors \(\langle 1,0,1 \rangle\) and \(\langle 0,2,1 \rangle\), we take the cross product of these vectors: \(\mathbf{n} = \langle 1,0,1 \rangle \times \langle 0,2,1 \rangle = \langle (0)(1)-(1)(2), (1)(0)-(1)(1), (1)(2)-(0)(1) \rangle = \langle -2, -1, 2 \rangle\)
03

Equation of the plane parallel to the vectors passing through the point (1,2,3)

Now that we have found the normal vector of this plane, which is \(\langle -2, -1, 2 \rangle\), we can use the formula for the equation of a plane with a point and a normal vector as in Step 1. The point given is (1,2,3), and the normal vector is \(\langle -2, -1, 2 \rangle\). Plug them into the formula: \((-2)(x-1)+(-1)(y-2)+(2)(z-3)=0\) This simplifies to: \(-2x+y-2z+5=0\). The equations of the planes are: 1. For the plane passing through point \(P_{0}\) with normal vector \(\mathbf{n}\): \((n_x)(x-x_{0})+(n_y)(y-y_{0})+(n_z)(z-z_{0})=0\) 2. For the plane parallel to the vectors \(\langle 1,0,1 \rangle\) and \(\langle 0,2,1 \rangle\) and passing through the point (1,2,3): \(-2x+y-2z+5=0\)

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 Vector
The normal vector plays a fundamental role when determining the equation of a plane. In three-dimensional geometry, a plane can be uniquely defined if we know a point through which it passes and a vector that is perpendicular to the plane, known as the normal vector. The normal vector, often denoted as \(\mathbf{n}\), provides a crucial insight into the orientation of the plane in space.

If we have a normal vector \(\mathbf{n} = \langle n_{x}, n_{y}, n_{z} \rangle\), any vector lying on the plane will be perpendicular to \(\mathbf{n}\). This is why, when we calculate the equation of the plane using the point-normal form \(\mathbf{n} \. (\vec{r} - \vec{r_0}) = 0\), where \(\vec{r}\) is an arbitrary point on the plane, and \(\vec{r_0}\) is the known point, it ensures that our plane correctly orientates perpendicularly to the normal vector.
Cross Product
In the context of vectors, the cross product is an operation that takes two vectors and returns a third vector which is perpendicular to both of the original vectors. This resultant vector's magnitude is proportional to the area of the parallelogram that the two vectors span, giving us a sense of how 'spread out' the vectors are in relation to each other.

To find a normal vector for a plane that is parallel to two given vectors, we can compute their cross product. For example, given vectors \(\mathbf{a} = \langle a_{x}, a_{y}, a_{z} \rangle\) and \(\mathbf{b} = \langle b_{x}, b_{y}, b_{z} \rangle\), the cross product \(\mathbf{n} = \mathbf{a} \times \mathbf{b}\) is calculated as \(\mathbf{n} = \langle a_{y}b_{z} - a_{z}b_{y}, a_{z}b_{x} - a_{x}b_{z}, a_{x}b_{y} - a_{y}b_{x} \rangle\). This vector \(\mathbf{n}\) is then used as the normal vector for the plane equation, as it guarantees perpendicularly to both \(\mathbf{a}\) and \(\mathbf{b}\).
Vector Parallelism
Two vectors are considered parallel if they have the same or exact opposite directions. Technically, this means one is a scalar multiple of the other. This property is essential when analyzing planes because a plane can be described regarding its parallelism to vectors.

If two non-parallel, non-zero vectors lie on a plane, they define the plane's orientation. Any vector parallel to these two vectors will also lie on the plane, effectively letting us visualize the plane as the collection of all linear combinations of the two vectors. Thus, when giving the exercise to find a plane parallel to two vectors through a certain point, understanding vector parallelism allows us to know that these vectors dictate the direction of the plane but do not affect its position in space which is determined by the point through which the plane passes.

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

Suppose you follow the spiral path \(C: x=\cos t, y\) \(=\sin t, z=\) \(t,\) for \(t \geq 0,\) through the domain of the function \(w=f(x, y, z)=x y z /\left(z^{2}+1\right)\) a. Find \(w^{\prime}(t)\) along \(C\) b. Estimate the point \((x, y, z)\) on \(C\) at which \(w\) has its maximum value.

Identify and briefly describe the surfaces defined by the following equations. $$x^{2} / 4+y^{2}-2 x-10 y-z^{2}+41=0$$

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).

Let \(f(x, y)=0\) define \(y\) as a twice differentiable function of \(x\) a. Show that \(y^{\prime \prime}(x)=\frac{f_{x x} f_{y}^{2}-2 f_{x} f_{y} f_{x y}+f_{y y} f_{x}^{2}}{f_{y}^{3}}\) b. Verify part (a) using the function \(f(x, y)=x y-1\)

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). $$u(x, t)=10 e^{-t} \sin x$$

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.