Problem 5 - University Calculus: Early Transcendentals

Question

Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). $$\mathbf{u}=2 \mathbf{i}, \quad \mathbf{v}=-3 \mathbf{j}$$

Step-by-Step Solution

Answer

\(\text{Length of } \mathbf{u} \times \mathbf{v} = 6\), direction: negative \(\mathbf{k}\); \(\text{length of } \mathbf{v} \times \mathbf{u} = 6\), direction: positive \(\mathbf{k}\).

Step 1: Understand the Problem

We need to find the cross product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\), as well as the direction of these cross products. The vectors are given as \(\mathbf{u}=2 \mathbf{i}\) and \(\mathbf{v}=-3 \mathbf{j}\).

Step 2: Cross Product Formula

The cross product \(\mathbf{a} \times \mathbf{b}\) is calculated using the determinant of a matrix formed by the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and the components of \(\mathbf{a}\) and \(\mathbf{b}\). If \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\), then:\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}.\]In our case, \(\mathbf{u} = 2 \mathbf{i}\) (\(a_1 = 2, a_2 = 0, a_3 = 0\)) and \(\mathbf{v} = -3 \mathbf{j}\) (\(b_1 = 0, b_2 = -3, b_3 = 0\)).

Step 3: Calculate \(\mathbf{u} \times \mathbf{v}\)

Using the cross product formula:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 0 & 0 \ 0 & -3 & 0 \end{vmatrix} = (0 - 0)\mathbf{i} - (0 - 0)\mathbf{j} + (2 \times (-3) - 0)\mathbf{k} = -6 \mathbf{k}.\]So, \(\mathbf{u} \times \mathbf{v} = -6 \mathbf{k}\).

Step 4: Determine Length of \(\mathbf{u} \times \mathbf{v}\)

The magnitude, or length, of a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) is \(\sqrt{a_1^2 + a_2^2 + a_3^2}\). Hence, the magnitude of \(-6 \mathbf{k}\) is \(\sqrt{0^2 + 0^2 + (-6)^2} = \sqrt{36} = 6\).

Step 5: Direction of \(\mathbf{u} \times \mathbf{v}\)

The direction of the cross product \(-6 \mathbf{k}\) is in the negative \(\mathbf{k}\) direction, which is along the negative z-axis.

Step 6: Calculate \(\mathbf{v} \times \mathbf{u}\)

The cross product \(\mathbf{v} \times \mathbf{u}\) is:\[ \mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -3 & 0 \ 2 & 0 & 0 \end{vmatrix} = (0 - 0)\mathbf{i} - (0 - 0)\mathbf{j} + (0 - 2\times(-3))\mathbf{k} = 6 \mathbf{k}.\]So, \(\mathbf{v} \times \mathbf{u} = 6 \mathbf{k}\).

Step 7: Determine Length of \(\mathbf{v} \times \mathbf{u}\)

The magnitude is \(\sqrt{6^2} = 6\) for \(6 \mathbf{k}\).

Step 8: Direction of \(\mathbf{v} \times \mathbf{u}\)

The direction of the vector \(6 \mathbf{k}\) is along the positive z-axis, which is in the positive \(\mathbf{k}\) direction.

Key Concepts

Vector Magnitude

The vector magnitude, often referred to as the length of a vector, gives a sense of how long the vector is. Calculating the magnitude is simple once you know the vector's components. If you have a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\), its magnitude is found using the formula \(\sqrt{a_1^2 + a_2^2 + a_3^2}\).
This formula is derived from the Pythagorean theorem, which is used to calculate distances in a plane. In the context of three-dimensional vectors, it tells us the straight-line distance from the origin to the point defined by \( (a_1, a_2, a_3) \).
To illustrate, consider the cross product we derived in the exercise as \(-6 \mathbf{k}\) and \(6 \mathbf{k}\). Using the magnitude formula:

• For \(-6 \mathbf{k}\), the magnitude is \(\sqrt{0^2 + 0^2 + (-6)^2} = \sqrt{36} = 6\).
• Similarly, for \(6 \mathbf{k}\), the magnitude is \(\sqrt{6^2} = \sqrt{36} = 6\).

Regardless of the sign of a component, magnitude is always a positive real number, representing the length of the vector.

Three-Dimensional Vectors

Three-dimensional vectors are an extension of the two-dimensional vectors you might already be familiar with. They give us the ability to describe motion and direction in 3D space. This includes the x, y, and z-axes, typically associated with the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) respectively.
A vector in three dimensions is expressed as \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), where \( a, b, \) and \( c \) are the scalar components of the vector. These components tell us how far the vector extends in each direction along the x, y, and z dimensions.

• In our exercise, \(\mathbf{u} = 2 \mathbf{i}\) is purely along the x-axis, meaning it has no components in the y and z directions.
• The vector \(\mathbf{v} = -3 \mathbf{j}\) only has a component along the y-axis, emphasizing its extension in that direction alone.

Understanding the components of these vectors in relation to each axis helps you visualize their combined effect when performing operations like the cross product.

Right-Hand Rule

The right-hand rule is a common technique used to determine the direction of the cross product of two vectors in three-dimensional space. This rule is essential because it helps visualize what cannot be easily seen through calculations alone.
To use the right-hand rule, you begin by pointing your right-hand fingers in the direction of the first vector (\(\mathbf{u}\)) and then curl them towards the second vector (\(\mathbf{v}\)). Your thumb will then point in the direction of the resulting cross product \(\mathbf{u} \times \mathbf{v}\).

• Applying this to the exercise, when we cross \(\mathbf{u} = 2 \mathbf{i}\) and \(\mathbf{v} = -3 \mathbf{j}\), using the right-hand rule guides us to the negative z-direction, which corresponds to the \(-6 \mathbf{k}\) outcome.
• The reverse operation, \(\mathbf{v} \times \mathbf{u}\), conversely points your thumb in the positive z-direction, yielding \(6 \mathbf{k}\).

The right-hand rule not only helps in determining direction but also emphasizes the anti-commutative property of the cross product. This means \( \mathbf{u} \times \mathbf{v} eq \mathbf{v} \times \mathbf{u} \); instead, they are negatives of each other.