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Simplify \(\|\vec{u} \times \vec{v}\|^{2}+(\vec{u} \bullet \vec{v})^{2}-\|\vec{u}\|^{2}\|\vec{v}\|^{2}\)

Short Answer

Expert verified
The simplified expression is 0.

Step by step solution

01

- Understand Vector Operations

Recall the definitions of the dot product and cross product: - The dot product of two vectors \(\vec{u} \bullet \vec{v}\) is a scalar and is given by \(\|\vec{u}||\vec{v}|\cos\theta\). - The magnitude of the cross product \(\|\vec{u} \times \vec{v}\|\) is given by \(\|\vec{u}\|\|\vec{v}\|\sin\theta\), where \(\theta\) is the angle between the vectors.
02

- Square the Cross Product Magnitude

Calculate the square of the magnitude of the cross product: \[\|\vec{u} \times \vec{v}\|^2 = (\|\vec{u}\|\|\vec{v}\|\sin\theta)^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 \sin^2\theta\].
03

- Square the Dot Product

Calculate the square of the dot product: \[ (\vec{u} \bullet \vec{v})^2 = (\|\vec{u}\|\|\vec{v}\|\cos\theta)^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 \cos^2\theta\].
04

- Combine Terms

Add the squared terms obtained from steps 2 and 3: \[\|\vec{u} \times \vec{v}\|^2 + (\vec{u} \bullet \vec{v})^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 \sin^2\theta + \|\vec{u}\|^2 \|\vec{v}\|^2 \cos^2\theta\].
05

- Simplify Using Trigonometric Identity

Use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\): \[ \|\vec{u} \times \vec{v}\|^2 + (\vec{u} \bullet \vec{v})^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 (\sin^2\theta + \cos^2\theta) = \|\vec{u}\|^2 \|\vec{v}\|^2 \].
06

- Subtract Given Term

Subtract \(\|\vec{u}\|^2 \|\vec{v}\|^2\) from the combined terms: \[\|\vec{u} \times \vec{v}\|^2 + (\vec{u} \bullet \vec{v})^2 - \|\vec{u}\|^2 \|\vec{v}\|^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 - \|\vec{u}\|^2 \|\vec{v}\|^2 = 0 \].

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Key Concepts

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

Dot Product
The dot product of two vectors, represented as \(\text{\(\vec{u} \bullet \vec{v}\)}\), is a measure of their alignment. It results in a scalar value and can be calculated using the formula \(\text{\(\vec{u} \bullet \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos \theta\)}\). Here, \(\text{\(\|\vec{u}\|\)}\) and \(\text{\(\|\vec{v}\|\)}\) are the magnitudes of the vectors, and \(\text{\(\theta\)}\) is the angle between them.

Key Properties of Dot Product:
  • It is commutative: \(\text{\(\vec{u} \bullet \vec{v} = \vec{v} \bullet \vec{u}\)}\).
  • If the vectors are perpendicular (orthogonal), then their dot product is zero: \(\text{\(\vec{u} \bullet \vec{v} = 0\)}\).
Dot products are incredibly useful in physics and engineering, especially when determining the angle between vectors or projecting one vector onto another.

In the context of the exercise, we squared the result of the dot product to reach our final simplification.
Cross Product
The cross product of two vectors, denoted as \(\text{\(\vec{u} \times \vec{v}\)}\), results in a vector that is perpendicular to both original vectors. The magnitude of the cross product is given by \(\text{\(\|\vec{u} \times \vec{v}\| = \|\vec{u}\| \|\vec{v}\| \sin \theta\)}\).

Main Features of Cross Product:
  • The result is a vector, not a scalar.
  • The direction of this new vector is determined by the right-hand rule.
  • The magnitude reflects the area of the parallelogram formed by the original vectors.
In our exercise, we squared the magnitude of the cross product as part of the simplification process.

This step was essential to utilizing trigonometric identities, leading to the final formula.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for any angle. The most relevant identity for this exercise is the Pythagorean identity: \(\text{\(\sin^2 \theta + \cos^2 \theta = 1\)}\). This identity is fundamental in simplifying expressions involving trigonometric functions.

Important Usages:
  • It helps convert between \(\text{\(\sin\)}\) and \(\text{\(\cos\)}\) functions.
  • It is often used in integrals, derivatives, and other algebraic manipulations.
In the provided solution, this identity allowed us to combine the squared sine and cosine terms effectively, leading to further simplification. Understanding and applying these identities is crucial for solving various problems in linear algebra and physics.
Vector Magnitude
The magnitude of a vector, represented as \(\text{\(\|\vec{u}\|\)}\), provides a measure of its length. It is calculated using the Euclidean norm: \(\text{\(\|\vec{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2}\)}\) for a vector \(\text{\(\vec{u} = (u_1, u_2, u_3)\)}\).

Essential Properties:
  • Magnitudes are always non-negative.
  • A vector with a magnitude of zero is called the zero vector.
In the exercise, we used the magnitudes of vectors to express both the dot and cross products terms. This helped in eventually simplifying the expression to zero.

Grasping vector magnitude is foundational to understanding more complex operations like dot products and cross products.

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Most popular questions from this chapter

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $$ \left[\begin{array}{r} 1 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{r} 2 \\ 1 \\ -1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right] $$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

As shown in the above problem, one can find the closest point to \(\vec{v}\) in a plane through the origin by finding the intersection of the line through \(\vec{v}\) having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by \(\vec{v} .\) Here is a relation which defines a plane $$ 2 x+y+z=11 $$ and here is a point: \((1,1,2) .\) Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points. Hint: Line: \((x, y, z)=(1,1,2)+t(2,1,1) .\) Now require that it intersect the plane.

The total force acting on an object is to be \(4 \vec{i}+2 \vec{j}-3 \vec{k}\) Newtons. A force of \(-3 \vec{i}-1 \vec{j}+8 \vec{k}\) Newtons is being applied. What other force should be applied to achieve the desired total force?

An object moves 10 meters in the direction of \(\vec{j} .\) There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+\vec{j}+2 \vec{k},\) and \(\vec{F}_{2}=-5 \vec{i}+2 \vec{j}-6 \vec{k}\). Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

The wind blows from West to East at a speed of 50 miles per hour and an airplane which travels at 400 miles per hour in still air heading somewhat West of North so that, with the wind, it is flying due North. It uses 30.0 gallons of gas every hour. If it has to travel 600.0 miles due North, how much gas will it use in flying to its destination?

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