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Chapter 4: Problem 17

In Exercises 13-18, find a \(\times\) b. \(a=2 i-3 j+5 k, b=4 i-5 j+k\)

Short Answer

Expert verified
The cross product \(a \times b\) is \(22i + 18j + 2k\).

Step by step solution

01

Identify the components of vectors

Write down the components of vectors \(a\) and \(b\) from the given equations. For vector \(a\), the components are \(a_1 = 2\), \(a_2 = -3\), and \(a_3 = 5\).For vector \(b\), the components are \(b_1 = 4\), \(b_2 = -5\), and \(b_3 = 1\).
02

Understand the cross product formula

The cross product \(a \times b\) of two vectors \(a = a_1 i + a_2 j + a_3 k\) and \(b = b_1 i + b_2 j + b_3 k\) is given by:\[ a \times b = (a_2b_3 - a_3b_2)i - (a_1b_3 - a_3b_1)j + (a_1b_2 - a_2b_1)k \] This formula involves calculating the determinant of a 3x3 matrix whose first row is \(i, j, k\), second row is the components of \(a\), and third row is the components of \(b\).
03

Calculate the i-component

Using the formula, the i-component of the cross product is calculated as:\[ i: a_2b_3 - a_3b_2 = (-3)(1) - (5)(-5) = -3 + 25 = 22 \].
04

Calculate the j-component

The j-component of the cross product is calculated as:\[ j: -(a_1b_3 - a_3b_1) = -(2 \times 1 - 5 \times 4) = -(2 - 20) = 18 \].
05

Calculate the k-component

The k-component of the cross product is calculated as:\[ k: a_1b_2 - a_2b_1 = 2(-5) - (-3)(4) = -10 + 12 = 2 \].
06

Write down the final cross product

Combine the components calculated in the previous steps to write down the cross product:\[ a \times b = 22i + 18j + 2k \].

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Key Concepts

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

Vector Components
Vectors in three-dimensional space consist of three parts: the i, j, and k components. These components represent the contributions of the vector along the x, y, and z axes, respectively.
For example, the vector \( a = 2i - 3j + 5k \) comprises the following components:
  • \( a_1 = 2 \) for the i or x-direction,
  • \( a_2 = -3 \) for the j or y-direction,
  • \( a_3 = 5 \) for the k or z-direction.
Similarly, for vector \( b = 4i - 5j + k \), its components are
  • \( b_1 = 4 \),
  • \( b_2 = -5 \),
  • \( b_3 = 1 \).
Understanding these components is crucial, as they are used in various vector operations such as calculating the cross product.
Determinant
The determinant is a special number calculated from a square matrix, often used to solve systems of linear equations, among other applications.
In the context of cross products, the determinant helps find the perpendicular vector resulting from the vector product.

For our exercise, to compute the cross product \( a \times b \), you can represent it using a 3x3 matrix:
  • The first row consists of unit vectors: \( i, j, \text{and} k \).
  • The second row has components of vector \( a \): \( a_1, a_2, \text{and} a_3 \).
  • The third row lists components of vector \( b \): \( b_1, b_2, \text{and} b_3 \).
The determinant of this matrix gives the cross product, providing a vector orthogonal to both \( a \) and \( b \). By arranging and solving this matrix, one finds the components \( 22i + 18j + 2k \).
3D Vectors
3D vectors extend the concept of two-dimensional vectors by adding a third component, which aligns with the z-axis, in addition to the x and y components.
This opens up applications in physics and engineering where phenomena exist in three dimensions.

These vectors are particularly useful in representing physical quantities such as velocity, force, and displacement in 3D space. For instance:
  • The vector \( a = 2i - 3j + 5k \) indicates movement with a forward and upward force magnitude in 3D space.
  • Their manipulation through vector operations like cross products can lead to insightful discoveries about motion, fields, or moment generation.
So, understanding cross products' computational methods and physical implications is vital in comprehending complex multi-dimensional systems.
Calculating Cross Products
The cross product of two vectors results in a vector that is perpendicular to both original vectors. This vector is often used for calculating torque, rotational motion, or determining perpendicular planes.
The calculation entails:
  • Forming a 3x3 matrix with \( i, j, k \) as the first row, the components of the first vector, and the components of the second vector as subsequent rows.
From our example:
  • The i-component: \( (-3)(1) - (5)(-5) = 22 \)
  • The j-component: \( -(2 \times 1 - 5 \times 4) = 18 \)
  • The k-component: \( 2(-5) - (-3)(4) = 2 \)
Segregating these, we combine them to yield the result: \( 22i + 18j + 2k \).

This result is not just an arithmetic expression, but a direction and magnitude expressing important spatial insights in 3D systems.

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Most popular questions from this chapter

Let \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be a linear transformation of rank \(n\) with standard matrix representation A. Mark each of the following True or False. a. The image under \(T\) of a box in \(\mathbb{R}^{n}\) is again a box in \(\mathbb{R}^{n}\). b. The image under \(T\) of an \(n\)-box in \(\mathbb{R}^{n}\) of volume \(V\) is a box in \(\mathbb{R}^{n}\) of volume \(\operatorname{det}(A) \cdot V\). c. The image under \(T\) of an \(n\)-box in \(\mathbb{R}^{n}\) of volume \(>0\) is a box in \(\mathbb{R}^{n}\) of volume \(>0\). d. If the image under \(T\) of an \(n\)-box \(B\) in \(\mathrm{R}^{n}\) has volume 12 , the box \(B\) has volume \(|\operatorname{det}(A)| \cdot 12\) If the image under \(T\) of an \(n\)-box \(B\) in \(\mathbb{R}^{n}\) has volume 12 , the box \(B\) has volume \(12 /|\operatorname{det}(A)|\). f. If \(n=2\), the image under \(T\) of the unit \(\operatorname{disk} x^{2}+y^{2} \leq 1\) has area \(|\operatorname{det}(A)|\). g. The linear transformation \(T\) is an isomorphism. h. The image under \(T \circ T\) of an \(n\)-box in \(\mathbb{R}^{n}\) of volume \(V\) is a box in \(\mathbb{R}^{n}\) of volume \(\operatorname{det}\left(A^{2}\right) \cdot V\). i. The image under \(T \circ T \circ T\) of an \(n\)-box in \(\mathbb{R}^{n}\) of volume \(V\) is a box in \(\mathbb{R}^{n}\) of volume \(\operatorname{det}\left(A^{3}\right) \cdot V\). j. The image under \(T\) of a nondegenerate 1-box is again nondegenerate.

Find the volume of the \(n \cdot b o x\) determined by the given vectors in \(R^{n}\). $$ \begin{aligned} &{[1,-1,0,1],[2,-1,3,1],[-1,4,2,-1],} \\ &{[0,1,0,2] \text { in } \mathbb{R}^{4}} \end{aligned} $$

In Exercises 41-44, find the volume of the ietrahedron having the given vertices. (Consider how the volume of a tetrahedron having three vectors from one point as edges is related to the volume of the box having the same three vectors as adjacent edges.J \((-3,0,1),(4,2,1),(0,1,7),(1,1,1)\)

Find the determinant of the given matrix. $$ \left[\begin{array}{rrrr} 2 & -1 & 0 & 0 \\ 4 & 5 & 0 & 0 \\ 0 & 0 & 3 & 6 \\ 0 & 0 & -4 & 2 \end{array}\right] $$

Let \(T: \mathbb{R}^{3} \rightarrow R^{3}\) be defined by \(T([x, y, z])=[x-2 y, 3 x+z, 4 x+3 y]\). Find the volume of the image under \(T\) of each of the given regions in \(\mathbf{R}^{3}\). The box \(0 \leq x \leq 2,-1 \leq y \leq 3,2 \leq z \leq 5\)
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