/*! 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 212 Consider points \(A(2,-3,4), B(0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 212

Consider points \(A(2,-3,4), B(0,1,2),\) and \(C(-1,2,0)\) a. Find the area of parallelogram \(A B C D\) with adjacent sides \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\). b. Find the area of triangle \(A B C\). c. Find the distance from point \(B\) to line \(A C\).

Short Answer

Expert verified
a. Area of parallelogram is \(2\sqrt{11}\). b. Area of triangle is \(\sqrt{11}\). c. Distance from B to AC is approximately 7.27.

Step by step solution

01

Calculate Vectors for Sides AB and AC

To find the vectors \( \overrightarrow{A B} \) and \( \overrightarrow{A C} \), we use the coordinates of points \( A \), \( B \), and \( C \).\- \( \overrightarrow{A B} = B - A = (0 - 2, 1 + 3, 2 - 4) = (-2, 4, -2) \)\- \( \overrightarrow{A C} = C - A = (-1 - 2, 2 + 3, 0 - 4) = (-3, 5, -4) \)
02

Use Cross Product to Find Parallelogram Area

The area of parallelogram is given by the magnitude of the cross product of \( \overrightarrow{A B} \) and \( \overrightarrow{A C} \). Calculate the cross product \( \overrightarrow{A B} \times \overrightarrow{A C} \):\\[\overrightarrow{A B} \times \overrightarrow{A C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 4 & -2 \ -3 & 5 & -4 \end{vmatrix}\]\= \( \mathbf{i}(4(-4) - (-2)(5)) - \mathbf{j}(-2(-4) - (-2)(-3)) + \mathbf{k}(-2*5 - 4*(-3)) \)\= \( \mathbf{i}(-16 + 10) - \mathbf{j}(8 - 6) + \mathbf{k}(-10 + 12) \)\= \( -6\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \)\The magnitude is \( \sqrt{(-6)^2 + (-2)^2 + 2^2} = \sqrt{36 + 4 + 4} = \sqrt{44} = 2\sqrt{11} \).\Thus, the area of the parallelogram \( ABCD \) is \( 2\sqrt{11} \).
03

Find Triangle Area Using Parallelogram Area

The area of triangle \(ABC\) is half of the area of parallelogram \(ABCD\).\\(\text{Area of } \triangle ABC = \frac{1}{2} \times 2 \sqrt{11} = \sqrt{11}\)
04

Find Line Equation for AC

The line for \( AC \) can be parameterized using point \( A (2, -3, 4) \) and direction vector \( \overrightarrow{AC} = (-3, 5, -4) \).\Line equation: \( (x, y, z) = (2, -3, 4) + t(-3, 5, -4) \)\Simplifying, \( x = 2 - 3t, y = -3 + 5t, z = 4 - 4t \).
05

Find projection of vector on line

Find vector \( \overrightarrow{B}A = A - B = (2, -3, 4) - (0, 1, 2) = (2, -4, 2) \).Project \( \overrightarrow{B}A \) on direction vector \( \overrightarrow{AC} \).\\( \text{Projection} = \left( \frac{\overrightarrow{BA} \cdot \overrightarrow{AC}}{\|\overrightarrow{AC}\|^2} \right) \overrightarrow{AC} \)\( \overrightarrow{BA} \cdot \overrightarrow{AC} = 2(-3) + (-4)5 + 2(-4) = -6 - 20 - 8 = -34 \)\( \| \overrightarrow{AC} \|^2 = (-3)^2 + 5^2 + (-4)^2 = 9 + 25 + 16 = 50 \)\( \text{Projection} = \left( \frac{-34}{50} \right)(-3, 5, -4) = (-1.02, 1.7, 1.36) \)
06

Calculate Distance from Point to Line

Calculate the perpendicular distance using vector subtraction. \\( \text{Distance Vector} = \overrightarrow{B}A - \text{Projection} \)\( = (2 - (-1.02), -4 - 1.7, 2 - 1.36 ) \)\( = (3.02, -5.7, 3.36) \)Calculate magnitude \( \sqrt{3.02^2 + (-5.7)^2 + 3.36^2} = \sqrt{9.1204 + 32.49 + 11.2896} = \sqrt{52.9} \approx 7.27 \).

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.

Cross Product
The cross product is a fundamental operation in vector calculus, crucial for finding the perpendicular vector to two given vectors in a three-dimensional space. Unlike the dot product, which results in a scalar, the cross product results in a vector. This vector is orthogonal to the plane formed by the original vectors, assuming they are not parallel.

To compute the cross product of two vectors, like \( \overrightarrow{A B} = (-2, 4, -2) \) and \( \overrightarrow{A C} = (-3, 5, -4) \), we use the determinant method:
  • Create a 3x3 matrix with unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) in the first row, components of the first vector in the second row, and components of the second vector in the third row.
  • Compute the determinant of this matrix to find each component of the cross product: \[ \overrightarrow{A B} \times \overrightarrow{A C} = \mathbf{i}(4(-4) - (-2)(5)) - \mathbf{j}(-2(-4) - (-2)(-3)) + \mathbf{k}(-2\cdot5 - 4(-3)) \]
  • This gives us the vector \((-6, -2, 2)\).
The magnitude of this vector represents the area of the parallelogram formed by the vectors.
Parallelogram Area
In vector calculus, the area of a parallelogram spanned by two vectors is found using the magnitude of their cross product. This method is both intuitive and efficient.

For vectors \( \overrightarrow{A B} = (-2, 4, -2) \) and \( \overrightarrow{A C} = (-3, 5, -4) \), the cross product yields the vector \((-6, -2, 2)\). To find the area, calculate the magnitude of this resulting vector:
  • Compute: \( \sqrt{(-6)^2 + (-2)^2 + 2^2} = \sqrt{36 + 4 + 4} = \sqrt{44} \)
  • The area of the parallelogram is \( 2\sqrt{11} \).
This approach underscores the geometric representation of vectors and is widely used in physics and engineering contexts.
Triangle Area
Finding the area of a triangle in three dimensions using vectors involves a straightforward application of the parallelogram area. Since a triangle can be seen as half of a parallelogram formed by the same adjacent sides, its area calculation uses this relationship directly.

Given the parallelogram area \( 2\sqrt{11} \), the area of triangle \( \triangle ABC \) is simply:
  • Divide the parallelogram鈥檚 area by 2: \( \frac{1}{2} \times 2\sqrt{11} = \sqrt{11} \).
This method is efficient, especially when a triangle is defined by vectors, and it highlights the cohesive nature of vector applications across different geometries.
Distance from Point to Line
Calculating the distance from a point to a line in 3D space involves projections and perpendicular distances, with vectors serving as the fundamental underpinning.

To determine this distance, consider:
  • Point \( B \) and line \( AC \), parameterized as \( (x, y, z) = (2, -3, 4) + t(-3, 5, -4) \).
  • Find vector \( \overrightarrow{BA} = (2, -4, 2) \).
  • Project \( \overrightarrow{BA} \) onto \( \overrightarrow{AC} \):
    \( \text{Projection} = \left( \frac{\overrightarrow{BA} \cdot \overrightarrow{AC}}{\|\overrightarrow{AC}\|^2} \right) \overrightarrow{AC} = (-1.02, 1.7, 1.36) \)
  • Calculate the actual distance vector by subtracting the projection from \( \overrightarrow{BA} \) then determine its magnitude:
  • Distance vector: \( (3.02, -5.7, 3.36) \); Magnitude: \( \sqrt{52.9} \approx 7.27 \).
This method elegantly combines geometry and algebra, showcasing vector calculus's power in practical applications.

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

For the following exercises, the equation of a surface in cylindrical coordinates is given. Find the equation of the surface in rectangular coordinates. Identify and graph the surface. [T] \(r=4\)

[T] A spheroid is an ellipsoid with two equal semiaxes. For instance, the equation of a spheroid with the \(z\) -axis as its axis of symmetry is given by \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1, \quad\) where \(a\) and \(c\) are positive real numbers. The spheroid is called oblate if \(ca\). a. The eye cornea is approximated as a prolate spheroid with an axis that is the eye, where \(a=8.7 \mathrm{~mm}\) and \(c=9.6 \mathrm{~mm} .\) Write the equation of the spheroid that models the cornea and sketch the surface. b. Give two examples of objects with prolate spheroid shapes.

For the following exercises, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in spherical coordinates. Identify the surface. \(x^{2}+y^{2}=9\)

Use the following figure as an aid in identifying the relationship between the rectangular, cylindrical, and spherical coordinate systems. For the following exercises, the cylindrical coordinates \((r, \theta, z)\) of a point are given. Find the rectangular coordinates \((x, y, z)\) of the point. \(\left(4, \frac{\pi}{6}, 3\right)\)

For the following exercises, the equation of a surface in spherical coordinates is given. Find the equation of the surface in rectangular coordinates. Identify and graph the surface. [T] \(\rho=3\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

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.