Chapter 10: Problem 42
Let \(\mathbf{v}=5 \mathbf{j}\) and let \(\mathbf{u}\) be a vector with length 3 that starts at the origin and rotates in the \(x y\) -plane. Find the maximum and minimum values of the length of the vector \(\mathbf{u} \times \mathbf{v} .\) In what direction does \(\mathbf{u} \times \mathbf{v}\) point?
Short Answer
Expert verified
The maximum magnitude of \(|\mathbf{u} \times \mathbf{v}|\) is 15, the minimum is 0. The vector \(\mathbf{u} \times \mathbf{v}\) points along the \(z\)-axis.
Step by step solution
01
Understanding the Vector Setup
We are given a vector \(\mathbf{v} = 5 \mathbf{j}\), which means \(\mathbf{v}\) points in the positive \(y\)-direction and has a magnitude of 5. \(\mathbf{u}\) is another vector in the \(xy\)-plane originating from the origin and has a constant magnitude of 3.
02
Expressing Vector \(\mathbf{u}\)
Since \(\mathbf{u}\) is a vector in the \(xy\)-plane that rotates, it can be expressed in terms of an angle \(\theta\) as \(\mathbf{u} = 3\cos(\theta) \mathbf{i} + 3\sin(\theta) \mathbf{j}\).
03
Calculating the Cross Product \(\mathbf{u} \times \mathbf{v}\)
The cross product of two vectors in three dimensions, \(\mathbf{u}\) and \(\mathbf{v}\), can be computed as follows: - Since \(\mathbf{v} = 0 \mathbf{i} + 5 \mathbf{j} + 0 \mathbf{k}\),- \(\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3\cos(\theta) & 3\sin(\theta) & 0 \ 0 & 5 & 0 \end{vmatrix}\).- Expanding the determinant gives \(\mathbf{u} \times \mathbf{v} = (0)\mathbf{i} - (0)\mathbf{j} + (15\cos(\theta))\mathbf{k} = 15\cos(\theta) \mathbf{k}\).
04
Finding the Magnitude of \(\mathbf{u} \times \mathbf{v}\)
The magnitude of the cross product vector is \(|\mathbf{u} \times \mathbf{v}| = |15\cos(\theta)|\). Since \(\cos(\theta)\) can vary between -1 and 1, the magnitude can vary between 0 and 15.
05
Determining Maximum and Minimum Values
The maximum value of \(|\mathbf{u} \times \mathbf{v}|\) is 15, when \(\cos(\theta) = 1\). The minimum value is 0, when \(\cos(\theta) = 0\).
06
Finding the Direction of \(\mathbf{u} \times \mathbf{v}\)
The direction of \(\mathbf{u} \times \mathbf{v}\) is along the \(z\)-axis because the cross product results in a vector \(15\cos(\theta) \mathbf{k}\). This direction is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\) as it points in the \(z\)-axis direction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
The magnitude of a vector, often referred to as its length, is a measure of how long the vector is in space. It is the distance from the vector's initial point to its terminal point. For a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude is calculated using the formula \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \). This is essentially the three-dimensional version of the Pythagorean theorem.
In our problem, the vector \( \mathbf{v} = 5\mathbf{j} \) has a magnitude of 5 already given, because its only component is along the \( y \)-axis. For the rotating vector \( \mathbf{u} \), with formula \( 3\cos(\theta)\mathbf{i} + 3\sin(\theta)\mathbf{j} \), the magnitude remains constant at 3, since \( \cos^2(\theta) + \sin^2(\theta) = 1 \)
In our problem, the vector \( \mathbf{v} = 5\mathbf{j} \) has a magnitude of 5 already given, because its only component is along the \( y \)-axis. For the rotating vector \( \mathbf{u} \), with formula \( 3\cos(\theta)\mathbf{i} + 3\sin(\theta)\mathbf{j} \), the magnitude remains constant at 3, since \( \cos^2(\theta) + \sin^2(\theta) = 1 \)
- A vector's magnitude provides insight into its strength without considering direction.
- Magnitudes are always non-negative values.
- For effective problem solving, know that any scalar multiply affects only magnitude, not direction.
Vector Direction
The direction of a vector indicates the way in which it points in space. For two-dimensional vectors, direction is commonly specified by an angle with respect to a reference axis, typically the positive x-axis. When a vector is represented as \( a\mathbf{i} + b\mathbf{j} \), the angle can be found using \( \theta = \arctan\left(\frac{b}{a}\right) \).
In the exercise, vector \( \mathbf{v} = 5\mathbf{j} \) points directly along the positive y-axis, indicating it moves up vertically without any horizontal component. The vector \( \mathbf{u} \) has a direction that rotates freely in the xy-plane based on the angle \( \theta \). With each rotation, \( \mathbf{u} \) traces a circle with a radius equal to its magnitude, which is 3.
In the exercise, vector \( \mathbf{v} = 5\mathbf{j} \) points directly along the positive y-axis, indicating it moves up vertically without any horizontal component. The vector \( \mathbf{u} \) has a direction that rotates freely in the xy-plane based on the angle \( \theta \). With each rotation, \( \mathbf{u} \) traces a circle with a radius equal to its magnitude, which is 3.
- It is essential to detach direction from magnitude when analyzing a vector's influence.
- In cross products, the resulting vector direction is determined by the respective orientations of the original vectors involved.
- The direction is vital in determining the vector's alignment and angle relative to coordinate axes.
Determinant Expansion
Determinant expansion is a technique used in calculus and linear algebra to compute the cross product of two vectors. For vector cross product computation, a 3x3 matrix is formed and expanded to find a resultant vector perpendicular to the original vectors.
The matrix layout is:
\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \3\cos(\theta) & 3\sin(\theta) & 0 \0 & 5 & 0 \end{vmatrix}\]
Using cofactor expansion along the first row, the expansion leads to:
* \( (0)\mathbf{i} - (0)\mathbf{j} + (15\cos(\theta))\mathbf{k} \), resulting in the simplified vector pointing purely in the \( z \)-axis
The matrix layout is:
\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \3\cos(\theta) & 3\sin(\theta) & 0 \0 & 5 & 0 \end{vmatrix}\]
Using cofactor expansion along the first row, the expansion leads to:
* \( (0)\mathbf{i} - (0)\mathbf{j} + (15\cos(\theta))\mathbf{k} \), resulting in the simplified vector pointing purely in the \( z \)-axis
- This method affirms that the cross product vector is orthogonal to both input vectors.
- The scalar results from the determinant can be positive, negative, or zero, directly affecting the magnitude of the cross product.
- This determinant expansion highlights the spatial relationship between vectors often used in physics to compute torques and rotational forces.