/*! 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 21 Sketch the plane parallel to the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the plane parallel to the \(x y\) -plane through (2,4,2) and find its equation.

Short Answer

Expert verified
Answer: The equation of the plane is z = 2.

Step by step solution

01

Determine the normal vector

Since the plane is parallel to the xy-plane, the normal vector will be the z-axis, which is given as \((0, 0, 1)\).
02

Write the general form of a plane's equation using the normal vector

The general form of a plane's equation using the normal vector \((A,B,C)\) and a point \((x_0, y_0, z_0)\) on the plane is given by: \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\). Using the normal vector \((0, 0, 1)\), we get: \(0(x-x_0) + 0(y-y_0) + 1(z-z_0) = 0\)
03

Plug in the given point (2, 4, 2)

The given point is (2, 4, 2). Substitute these values into the equation we got from step 2: \(0(x-2) + 0(y-4) + 1(z-2) = 0\)
04

Simplify the equation

Simplify the equation to obtain the final form: \(z-2 = 0\) Thus, the equation of the plane parallel to the xy-plane through point (2, 4, 2) is \(z = 2\).

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!

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

An object moves along a straight line from the point \(P(1,2,4)\) to the point \(Q(-6,8,10)\) a. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with a constant speed over the time interval [0,5] b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{t}\)

Consider the trajectory given by the position function $$\mathbf{r}(t)=\left\langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5\left(1-e^{-t}\right)\right), \quad \text { for } t \geq 0$$ a. Find the initial point \((t=0)\) and the "terminal" point \(\left(\lim _{t \rightarrow \infty} \mathbf{r}(t)\right)\) of the trajectory. b. At what point on the trajectory is the speed the greatest? c. Graph the trajectory.

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{array}{l} \mathbf{r}(t)=\langle 4+5 t,-2 t, 1+3 t\rangle ;\\\ \mathbf{R}(s)=\langle 10 s, 6+4 s, 4+6 s\rangle \end{array}$$

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(-12,4) ; \ell: y=2 x\)

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.