/*! 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 23 A plane contains the points \(A(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 23

A plane contains the points \(A(-4,9,-9)\) and \(B(5,-9,6)\) and is perpendicular to the line which joins \(B\) and \(C(4,-6, \mathrm{k})\). Obtain \(k\) and the equation of the plane.

Short Answer

Expert verified
Answer: The value of k is 10.2 and the equation of the plane is -x + 3y + 4.2z + 6.8 = 0.

Step by step solution

01

Find the direction vector BC and AB

Let \(\vec{B} = (5, -9, 6)\) and \(\vec{C} = (4, -6, k)\). To find the direction vector \(\vec{BC}\), subtract \(\vec{B}\) from \(\vec{C}\). $$ \vec{BC} = \vec{C} - \vec{B} = (4-5, -6 - (-9), k-6) = (-1, 3, k-6) $$ Now let \(\vec{A} = (-4, 9, -9)\). To find direction vector \(\vec{AB}\), subtract \(\vec{A}\) from \(\vec{B}\), $$ \vec{AB} = \vec{B} - \vec{A} = (5 - (-4), -9 - 9, 6 - (-9)) = (9, -18, 15) $$
02

Compute the dot product of direction vectors BC and AB

As the plane is perpendicular to the line joining B and C, the direction vector BC is normal to the plane. So we need to find if the dot product of direction vectors BC and AB is zero. $$ \vec{BC} \cdot \vec{AB} = (-1)(9)+ (3)(-18) + (k-6)(15) = 0 $$
03

Solve for k

Now, solve the dot product equation for k: $$ -9 - 54 + 15k - 90 = 0 $$ Combine like terms and solve for k: $$ 15k -153 = 0 $$ Divide by 15: $$ k -10.2 = 0 $$ So, the value of k is 10.2.
04

Find the equation of the plane

The normal vector of the plane, \(\vec{n}\), is the direction vector of BC, which is \(\vec{n} = (-1, 3, k-6) = (-1, 3, 4.2)\). We can find the equation of the plane using the point-normal form: $$ (ax - x_0) + (by - y_0) + (cz - z_0) = 0 $$ Substitute the components of the normal vector and the coordinates of point A: $$ (-1)(x - (-4)) + (3)(y - 9) + (4.2)(z - (-9)) = 0 $$ Simplify the equation: $$ - x - 4+ 3y -27 + 4.2z +37.8 = 0 $$ Combine like terms: $$ -x + 3y + 4.2z + 6.8 = 0 $$ This is the equation of the plane. So, the value of k is 10.2 and the equation of the plane is \(-x + 3y + 4.2z + 6.8 = 0\).

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 91影视!

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

The plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) meets the coordinate axes in \(A, B\) and \(C\), respectively.

A plane contains the points \(A(-4,9,-9)\) and \(B(5,-9,6)\) and is perpendicular to the line which joins \(B\) and \(C(4,-6, \mathrm{k})\). Obtain \(k\) and the equation of the plane.

Find the equation of the plane bisecting the line joining the points \((2,3,-1)\) and \((-5,6,3)\) at right angles.

\(O A B C\) is a tetrahedron in which \(O A, O B\) and \(O C\) are mutually perpendicular. Prove that the perpendicular from \(O\) to the base \(A B C\) meets it at its orthocentre.

Show that the points \((0,-1,0),(2,1,-1),(1,1,1)\) and \((-3,3,0)\) are coplanar.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.