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Chapter 2: Problem 184

For the following exercises, the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are given. a. Find the cross product \(\mathbf{u} \times \mathbf{v}\) of the vectors \(\mathbf{u}\) and v. Express the answer in component form. b. Sketch the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u} \times \mathbf{v}\). \(\mathbf{u}=\langle 3,2,-1\rangle, \quad \mathbf{v}=\langle 1,1,0\rangle\)

Short Answer

Expert verified
The cross product is \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\).

Step by step solution

01

Identify the Formula for the Cross Product

The cross product \(\mathbf{u} \times \mathbf{v}\) of two vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) is given by the formula: \[\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, \, u_3v_1 - u_1v_3, \, u_1v_2 - u_2v_1 \rangle\] Let's use this formula to find the cross product of the given vectors.
02

Substitute the Values into the Formula

Substitute the components of \(\mathbf{u}\) and \(\mathbf{v}\) into the cross product formula:\[\mathbf{u} \times \mathbf{v} = \langle 2 \cdot 0 - (-1) \cdot 1, \, -1 \cdot 1 - 3 \cdot 0, \, 3 \cdot 1 - 2 \cdot 1 \rangle\]
03

Compute Each Component of the Cross Product

Calculate each component of the cross product:1. First component: \(2 \cdot 0 - (-1) \cdot 1 = 0 + 1 = 1\)2. Second component: \((-1) \cdot 1 - 3 \cdot 0 = -1 \cdot 1 = -1\)3. Third component: \(3 \cdot 1 - 2 \cdot 1 = 3 - 2 = 1\)Thus, \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\).
04

Sketch the Vectors

Draw vectors \(\mathbf{u} = \langle 3,2,-1 \rangle\) and \(\mathbf{v} = \langle 1,1,0 \rangle\) in 3-dimensional space. Use standard basis vectors \(\hat{i}, \hat{j}, \hat{k}\) for x, y, and z axes.After drawing \(\mathbf{u}\) and \(\mathbf{v}\), calculate and draw the cross product vector \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\), which will be perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\). In 3D space, visualize it using the right-hand rule.

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Key Concepts

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

Vector Mathematics
Vectors are fundamental entities in mathematics and physics used to represent quantities that have both magnitude and direction. This makes them uniquely powerful for describing anything that moves or has a position.
  • Components of a Vector: Vectors in 3-dimensional space are typically represented as a set of three components. For instance, the vector \( \mathbf{u} = \langle 3, 2, -1 \rangle \) describes movement or position along the x, y, and z axes.
  • Cross Product: The cross product is a mathematical operation that takes two vectors and produces a third vector, which is perpendicular to the original two. This operation is denoted by the symbol "脳". The result is distinct from the dot product because it's a vector, not a scalar.
  • Importance: In physics, the cross product is crucial for understanding rotational forces and moments, where the direction and magnitude of the force are critical.
For vectors \( \mathbf{u} \) and \( \mathbf{v} \), the cross product is calculated using the formula: \[\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, \ u_3v_1 - u_1v_3, \ u_1v_2 - u_2v_1 \rangle\].This formula helps compute the components of the new vector.
3-Dimensional Space
Three-dimensional space extends the plane geometry into a new, richer environment. Familiarizing yourself with the 3D coordinate system is key to understanding vector mathematics.
  • Coordinate Axes: The x, y, and z axes form the foundation of 3D space. These axes are mutually perpendicular to each other, creating a cube-like space for manipulation and visualization.
  • Sketching Vectors: When sketching vectors in 3D, it's crucial to visualize them as arrows originating from a common point, typically the origin. This helps in visually understanding their direction and relative position.
    For example, to depict \( \mathbf{u} = \langle 3, 2, -1 \rangle \), you move 3 units along x, 2 units along y, and -1 unit along z. The negative z-component means the vector points towards the negative z-axis.
  • Applications: Understanding 3D vectors is critical in fields like physics, computer graphics, and engineering, where real-world problems require modeling beyond flat surfaces.
By representing vectors in 3D, we grasp more complex spatial relationships and interactions, such as forces or trajectories in physics.
Right-Hand Rule
The right-hand rule is a mnemonic used for understanding the direction of the resulting vector in a cross product operation.
  • How to Use: To apply the right-hand rule, align your right hand with the first vector (\(\mathbf{u}\)). Then, curl your fingers toward the second vector (\(\mathbf{v}\)), keeping your thumb extended. Your thumb will point in the direction of \(\mathbf{u} \times \mathbf{v}\).
  • Why It Works: This rule provides an intuitive grasp on the orientation of vectors in space. It concretizes the abstract calculation, giving a physical sense to the perpendicularity resulting from the cross product.
  • Visualization: In 3D space, using the right-hand rule ensures clarity when visualizing vector orientations. For instance, after computing \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\), the right-hand rule confirms that this vector is indeed perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).
This approach is essential in fields such as electromagnetism and computer graphics, where orientation and directionality matter.

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Most popular questions from this chapter

For the following exercises, determine whether the statement is true or false. Justify the answer with a proof or a counterexample. For vectors \(\mathbf{a}\) and \(\mathbf{b}\) and any given scalar \(c\), \(c(\mathbf{a} \times \mathbf{b})=(c \mathbf{a}) \times \mathbf{b}\)

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For the following exercises, consider a small boat crossing a river. An airplane is flying in the direction of \(52^{\circ}\) east of north with a speed of \(450 \mathrm{mph}\). A strong wind has a bearing \(33^{\circ}\) east of north with a speed of \(50 \mathrm{mph}\). What is the resultant ground speed and bearing of the airplane?

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