/*! 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 1 The equation of the plane throug... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 1

The equation of the plane through the points \((2,3,1)\) and \((4,-5,3)\) and parallel to \(x\)-axis is (A) \(x-z-1=0\) (B) \(4 x+y-11=0\) (C) \(y+4 z-7=0\) (D) none of these

Short Answer

Expert verified
The correct answer is (C) \(y+4z-7=0\).

Step by step solution

01

Determine the direction vector

To find the equation of the plane, we first need a normal vector to the plane. Since the plane is parallel to the x-axis, the direction vector along the x-axis is given by \( \mathbf{d} = (1,0,0) \). We can determine the direction vector from the points \( (2,3,1) \) and \( (4,-5,3) \) as \( \mathbf{v} = (4-2, -5-3, 3-1) = (2, -8, 2) \).
02

Determine another vector parallel to the plane

The plane must have another vector parallel to it, that is orthogonal to the x-axis. We found this vector \( \mathbf{v} = (2, -8, 2) \).
03

Find the normal vector to the plane

The normal vector to the plane \( \mathbf{n} \) is found using the cross product of the direction vectors \( \mathbf{d} = (1,0,0) \) and \( \mathbf{v} = (2, -8, 2) \). The cross product is \[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 0 \ 2 & -8 & 2 \end{vmatrix} = (0, -2, -8) \].
04

Write the equation of the plane

The equation of the plane can be written using the point-normal form as \( \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \), where \( \mathbf{r_0} \) is any point on the plane, for example, \( (2,3,1) \), and \( \mathbf{n} = (0, -2, -8) \). Substituting, we have \( 0(x - 2) - 2(y - 3) - 8(z - 1) = 0 \).
05

Simplify the equation

Simplify the equation from Step 4: \( -2(y - 3) - 8(z - 1) = 0 \). Expanding gives \( -2y + 6 - 8z + 8 = 0 \). Simplifying further, we get \( -2y - 8z + 14 = 0 \), which can be rewritten as \( 2y + 8z = 14 \). Dividing through by 2 gives \( y + 4z = 7 \).
06

Match the simplified equation to the given options

The simplified equation \( y + 4z = 7 \) matches option (C) in the given question. Thus, the right choice is option (C).

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 91影视!

Key Concepts

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

Direction Vector
A direction vector is a vector that indicates the direction of a line or a vector. It is particularly helpful when finding the equation of a plane. In the context of the exercise, the direction vector can be determined by the difference in coordinates of two points on a line. We used the points \((2,3,1)\) and \((4,-5,3)\) to find the direction vector \(\mathbf{v} = (4-2, -5-3, 3-1)\). This simplifies to \(\mathbf{v} = (2, -8, 2)\).
  • The x-component is \(4-2=2\).
  • The y-component is \(-5-3=-8\).
  • The z-component is \(3-1=2\).
Direction vectors are essential for determining orientation and geometric relationships in planes.
Normal Vector
The normal vector is a perpendicular vector to a plane, critical in defining the plane equation. In our problem, the normal vector \(\mathbf{n}\) is derived through the cross product of two vectors that lie on the plane.
  • A normal vector is essential for uniquely defining a plane's position and orientation.
  • For the plane parallel to the x-axis, we started with the direction vector \(\mathbf{d} = (1,0,0)\).
The normal vector was computed as \(\mathbf{n} = (0, -2, -8)\), which forms the backbone for the plane equation.
Cross Product
The cross product is an operation used to find a vector that is perpendicular to two given vectors, often applied in physics and engineering. In this exercise, it's used to find the normal vector. With vectors \(\mathbf{d} = (1,0,0)\) and \(\mathbf{v} = (2,-8,2)\), their cross product is \[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 0 \ 2 & -8 & 2 \end{vmatrix} = (0, -2, -8) \].
  • The result is perpendicular to both input vectors.
  • The cross product helps derive the normal vector which is crucial for planes.
The use of cross products simplifies finding perpendicular relationships in geometry.
Point-Normal Form
The point-normal form is a method to express the equation of a plane, using a point on the plane and the normal vector. It is given by the formula \( \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \), where \(\mathbf{r_0}\) is a point on the plane and \(\mathbf{n}\) is the normal vector.
  • We use it to transform geometric information into a plane equation.
  • It ensures that the plane is accurately depicted through its orientation in 3D space.
Applying this, with \( \mathbf{r_0} = (2,3,1) \) and \(\mathbf{n} = (0, -2, -8)\), gives \(0(x-2) - 2(y-3) - 8(z-1) = 0\). This simplifies to the final equation \( y + 4z = 7 \), matching the supposed solution.

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 equation of a plane parallel to the plane \(x-2 y+2 z\) \(=5\) and at a unit distance from the origin is \([\mathbf{2 0 1 2}]\) (A) \(x-2 y+2 z-3=0\) (B) \(x-2 y+2 z+1=0\) (C) \(x-2 y+2 z-1=0\) (D) \(x-2 y+2 z+5=0\)

The distance of the point \((1,0,2)\) from the point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane \(x-y+z=16\), is: (A) 8 (B) \(3 \sqrt{21}\) (C) 13 (D) \(2 \sqrt{14}\)

A line makes the same angle \(\theta\), with each of the \(x\) and \(z\) axis. If the angle \(\beta\), which it makes with \(y\)-axis, is such that \(\sin ^{2} \beta=3 \sin ^{2} \theta\), then \(\cos ^{2} \theta\) equals (A) \(\frac{2}{3}\) (B) \(\frac{1}{5}\) (C) \(\frac{3}{5}\) (D) \(\frac{2}{5}\)

3 If the plane \(2 a x-3 a y+4 a z+6=0\) passes through the midpoint of the line joining the centres of the spheres \([\mathbf{2 0 0 5}]\) \(x^{2}+y^{2}+z^{2}+6 x-8 y-2 z=13\) and \(x^{2}+y^{2}+z^{2}-10 x+4 y-2 z=8\), then a equals (A) \(-1\) (B) 1 (C) \(-2\) (D) 2

A square \(A B C D\) of diagonal \(2 a\) is folded along the diagonal \(A C\) so that the planes \(D A C\) and \(B A C\) are at right angle. The shortest distance between \(D C\) and \(A B\) is (A) \(\sqrt{2} a\) (B) \(2 a / \sqrt{3}\) (C) \(2 a / \sqrt{5}\) (D) \((\sqrt{3} / 2) a\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
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.