Chapter 11: Problem 5
Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). $$\mathbf{u}=2 \mathbf{i}, \quad \mathbf{v}=-3 \mathbf{j}$$
Short Answer
Expert verified
\(\text{Length of } \mathbf{u} \times \mathbf{v} = 6\), direction: negative \(\mathbf{k}\); \(\text{length of } \mathbf{v} \times \mathbf{u} = 6\), direction: positive \(\mathbf{k}\).
Step by step solution
01
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 \(\mathbf{v}=-3 \mathbf{j}\).
02
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 \(\mathbf{v} = -3 \mathbf{j}\) (\(b_1 = 0, b_2 = -3, b_3 = 0\)).
03
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, \(\mathbf{u} \times \mathbf{v} = -6 \mathbf{k}\).
04
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 \(\sqrt{0^2 + 0^2 + (-6)^2} = \sqrt{36} = 6\).
05
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.
06
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, \(\mathbf{v} \times \mathbf{u} = 6 \mathbf{k}\).
07
Determine Length of \(\mathbf{v} \times \mathbf{u}\)
The magnitude is \(\sqrt{6^2} = 6\) for \(6 \mathbf{k}\).
08
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \(\sqrt{6^2} = \sqrt{36} = 6\).
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.
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.
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 \(\mathbf{u} \times \mathbf{v}\).
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 \(6 \mathbf{k}\).