Problem 4 - Thomas Calculus

Question

In Exercises \(1-8,\) find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u} .\) $$ \mathbf{u}=\mathbf{i}+\mathbf{j}-\mathbf{k}, \quad \mathbf{v}=\mathbf{0} $$

Step-by-Step Solution

Answer

The length is 0 and the direction is undefined for both \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \).

Step 1: Understand the Problem

We need to find the cross products \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \). In this problem, \( \mathbf{u} = \mathbf{i} + \mathbf{j} - \mathbf{k} \) and \( \mathbf{v} = \mathbf{0} \), the zero vector.

Step 2: Recall the Cross Product Property

The cross product of any vector with the zero vector is the zero vector itself. Hence, \( \mathbf{u} \times \mathbf{0} = \mathbf{0} \) and \( \mathbf{0} \times \mathbf{u} = \mathbf{0} \).

Step 3: Determine Length and Direction

The length, also known as the magnitude, of the zero vector \( \mathbf{0} \) is 0. Since the zero vector has no magnitude, the direction of \( \mathbf{0} \) is undefined.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus used to find a vector that is perpendicular to two given vectors in three-dimensional space. It is denoted by \( \mathbf{u} \times \mathbf{v} \) and is crucial in physics and engineering for tasks involving rotational forces and torque. To compute the cross product of vectors \( \mathbf{u} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{v} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), use the determinant of a matrix:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]Calculating the determinant involves expanding it to yield:- \( \mathbf{i}(a_2b_3 - a_3b_2) - \mathbf{j}(a_1b_3 - a_3b_1) + \mathbf{k}(a_1b_2 - a_2b_1) \) which gives a new vector representing the cross product.
Two special properties of the cross product make it particularly interesting:

• The cross product is anti-commutative, meaning \( \mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \).
• The cross product of any vector with the zero vector is always the zero vector itself.

This last property is shown in our exercise, where \( \mathbf{u} \times \mathbf{0} = \mathbf{0} \) and \( \mathbf{0} \times \mathbf{u} = \mathbf{0} \).
This concept juggles mathematical operation and geometric interpretation, making it a powerful tool.

Zero Vector

The zero vector, denoted by \( \mathbf{0} \), is a unique entity in vector calculus. It is defined as the vector with all its components equal to zero. In three-dimensional space, it can be represented as:\[ \mathbf{0} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} \]This vector serves as an identity element for vector addition and possesses key properties that simplify calculations:- Adding the zero vector to any vector \( \mathbf{a} \) yields \( \mathbf{a} \) itself. - Example: \( \mathbf{a} + \mathbf{0} = \mathbf{a} \)- Its magnitude is always zero, making the computation straightforward.In the context of the cross product, multiplying any vector with the zero vector results in the zero vector. This is illustrated in our exercise where both \( \mathbf{u} \times \mathbf{0} \) and \( \mathbf{0} \times \mathbf{u} \) yield \( \mathbf{0} \).
The simplicity of the zero vector is one of its vital characteristics, which makes it handy in solving equations and understanding basic properties of vector spaces. Unlike other vectors, the zero vector does not have a direction, as it has no magnitude to indicate any space orientation.

Magnitude and Direction

Magnitude and direction are essential attributes of any non-zero vector. The magnitude, given by \( |\mathbf{v}| \), is the length of the vector, computed using the Pythagorean theorem extended into space:For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is:\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]This value tells us how long the vector is, independent of its position in space.
The direction of a vector is expressed using direction angles or a unit vector pointing in the same direction as the original vector. To find the unit vector \( \hat{\mathbf{v}} \) of \( \mathbf{v} \), divide each component by the magnitude:\[ \hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|}(a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \]However, for the zero vector, these concepts become undefined due to the absence of magnitude and direction. In our exercise, because both cross products result in the zero vector \( \mathbf{0} \), we say that the magnitude is zero, and direction is not determinable. This highlights the limitation of handling zero vectors in terms of directionality.
Understanding these key concepts of magnitude and direction helps in visualizing vectors spatially and conduces the analysis of vector-related problems across disciplines.