/*! 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 46 Find an area vector for the para... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an area vector for the parallelogram with given vertices. $$\begin{aligned}&P=(-1,-2,0), Q=(0,-1,0), R=(-2,-4,1)\\\&S=(-1,-3,1)\end{aligned}$$

Short Answer

Expert verified
The area vector of the parallelogram is \( \hat{i} - \hat{j} - \hat{k} \).

Step by step solution

01

Understand Parallelogram's Property

A parallelogram on the coordinate plane can be defined by its vertices. An area vector of a parallelogram is found using the cross product of its diagonals or adjacent sides vector.
02

Determine Vectors of Adjacent Sides

Calculate the vectors \( \vec{PQ} \) and \( \vec{PS} \). \( \vec{PQ} = Q - P = (0 + 1, -1 + 2, 0 - 0) = (1, 1, 0) \) and \( \vec{PS} = S - P = (-1 + 1, -3 + 2, 1 - 0) = (0, -1, 1) \).
03

Calculate Cross Product

The area vector is the cross product of \( \vec{PQ} \) and \( \vec{PS} \). This can be calculated as \( \vec{PQ} \times \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 0 & -1 & 1 \end{vmatrix} = (1\cdot1 - 0\cdot(-1))\hat{i} - (1\cdot1 - 0\cdot0)\hat{j} + (1\cdot(-1) - 1\cdot0)\hat{k} = \hat{i} - \hat{j} - \hat{k} \).
04

Verify Orthogonality for Correctness

Verify if the cross product results in a vector that would be orthogonal to both \( \vec{PQ} \) and \( \vec{PS} \). Dot product \( (1, 1, 0) \cdot (1, -1, -1) = 1\cdot1 + 1\cdot(-1) + 0\cdot1 = 0 \) and \( (0, -1, 1) \cdot (1, -1, -1) = 0\cdot1 + (-1)\cdot(-1) + 1\cdot(-1) = 0 \), confirming orthogonality.

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Key Concepts

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

Cross Product
In vector calculus, the cross product is a critical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space. This operation is essential in physics and engineering because it helps define the orientation and area of certain geometric objects, like parallelograms.
The cross product of two vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \) can be computed using the determinant:
  • \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)
  • This results in the vector \( (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k} \)
This vector is always orthogonal to the original vectors \( \vec{a} \) and \( \vec{b} \). The magnitude of this cross product also gives the area of the parallelogram formed by \( \vec{a} \) and \( \vec{b} \).
Remember, the direction of the cross product is determined by the right-hand rule: if you point your index finger in the direction of \( \vec{a} \) and your middle finger in the direction of \( \vec{b} \), your thumb will point in the direction of the cross product.
Area Vector
An area vector represents both the magnitude and direction of a surface in three-dimensional space. In the context of a parallelogram, the area vector is particularly useful. It is not merely about the size of the surface but also the way it's oriented in space.
To find the area vector of a parallelogram defined by vectors \( \vec{a} \) and \( \vec{b} \), we calculate the cross product \( \vec{a} \times \vec{b} \). The direction of this vector is orthogonal (perpendicular) to the entire plane of the parallelogram. The magnitude of this area vector is equal to the actual area of the parallelogram. Therefore:
  • The length of \( \vec{a} \times \vec{b} \) gives the area of the surface.
  • The unit vector in the direction of the cross product gives the normal direction.
This means the cross product provides a succinct way to capture both the size and the orientation of the space a surface occupies.
Orthogonality
Orthogonality is a concept that describes how two vectors are perpendicular to each other. In simpler terms, it means that two vectors meet at right angles. This is a central idea in linear algebra and vector calculus.
To test for orthogonality, we use the dot product of the two vectors. If the dot product is zero, the vectors are orthogonal. Let's say you have vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \):
  • The dot product is calculated as \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
  • If the result is zero, \( \vec{a} \) and \( \vec{b} \) are orthogonal.
For example, if the area vector (found via cross product) is orthogonal to the sides of the parallelogram like \( \vec{PQ} \) and \( \vec{PS} \), it verifies the correctness of the solution, as seen in our exercise.
Orthogonality is not just limited to vectors in mathematics; it is a principle applied across various fields like computer graphics, physics, and engineering to ensure components are independently functioning or "uncoupled."

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Most popular questions from this chapter

Given \(\vec{v}=3 \vec{i}+4 \vec{j}\) and force vector \(\vec{F}\) find: (a) The component of \(\vec{F}\) parallel to \(\vec{v}\). (b) The component of \(\vec{F}\) perpendicular to \(\vec{v}\). (c) The work, \(W\), done by force \(\vec{F}\) through displacement \(\vec{v}\). $$\vec{F}=4 \vec{i}+\vec{j}$$

A street vendor sells six items, with prices \(p_{1}\) dollars per unit, \(p_{2}\) dollars per unit, and so on. The vendor's price vector is \(\vec{p}=\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)=\) \((1.00,3.50,4.00,2.75,5.00,3.00) .\) The vendor sells \(q_{1}\) units of the first item, \(q_{2}\) units of the second item, and so on. The vendor's quantity vector is \(\vec{q}=\) \(\left(q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}\right)=(43,57,12,78,20,35) .\) Find \(\vec{p}\) \(\vec{q},\) give its units, and explain its significance to the vendor.

Perform the following operations on the given 3 -dimensional vectors.$$\vec{a}=2 \vec{j}+\vec{k} \quad \vec{b}=-3 \vec{i}+5 \vec{j}+4 \vec{k} \quad \vec{c}=\vec{i}+6 \vec{j}$$ $$\vec{y}=4 \vec{i}-7 \vec{j} \quad \vec{z}=\vec{i}-3 \vec{j}-\vec{k}$$ $$\vec{a} \cdot(\vec{c}+\vec{y})$$

Find an equation of a plane that satisfies the given conditions. Through (-2,3,2) and parallel to \(3 x+y+z=4\)

Perform the following operations on the given 3 -dimensional vectors.$$\vec{a}=2 \vec{j}+\vec{k} \quad \vec{b}=-3 \vec{i}+5 \vec{j}+4 \vec{k} \quad \vec{c}=\vec{i}+6 \vec{j}$$ $$\vec{y}=4 \vec{i}-7 \vec{j} \quad \vec{z}=\vec{i}-3 \vec{j}-\vec{k}$$ $$\vec{c} \cdot \vec{y}$$

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