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True or false, by testing on \(\mathbf{A}=\mathbf{i}, \mathbf{B}=\mathbf{j}, \mathbf{C}=\mathbf{k}:\) (a) \(A \times(A \times B)=0\) (b) \(\mathbf{A} \cdot(\mathbf{B} \times \mathbf{C})=(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\) (c) \(A \cdot(B \times C)=C \cdot(B \times A)\) (d) \((\mathbf{A}-\mathbf{B}) \times(\mathbf{A}+\mathbf{B})=2(\mathbf{A} \times \mathbf{B})\)

Short Answer

Expert verified
(a) False, (b) True, (c) True, (d) True.

Step by step solution

01

Evaluate (a) using vector identities

For part (a), we need to compute \( \mathbf{i} \times (\mathbf{i} \times \mathbf{j}) \). First, calculate \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \). Then, calculate \( \mathbf{i} \times \mathbf{k} = -\mathbf{j} \). Since \(-\mathbf{j} eq 0\), the statement is false.
02

Evaluate (b) using scalar triple product

For part (b), compute both sides. The left side: \( \mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = \mathbf{i} \cdot \mathbf{i} = 1 \). The right side: \( (\mathbf{i} \times \mathbf{j}) \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{k} = 1 \). Both sides are equal, so the statement is true.
03

Evaluate (c) using scalar triple product symmetry

Part (c) involves checking if \( \mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = \mathbf{k} \cdot (\mathbf{j} \times \mathbf{i}) \). From step 2, \( \mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = 1 \). Also, \( B \times A = - (A \times B) \) means \( \mathbf{k} \cdot (\mathbf{k}) = 1 \). Thus, the statement is true.
04

Evaluate (d) using vector identities

For part (d), calculate both expressions. The left side: \( (\mathbf{i} - \mathbf{j}) \times (\mathbf{i} + \mathbf{j}) = \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{j} - \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{j} = \mathbf{k} + \mathbf{k} = 2\mathbf{k} \). The right side: \( 2(\mathbf{i} \times \mathbf{j}) = 2\mathbf{k} \). Both sides are equal, so the statement is true.

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Key Concepts

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

Vector Identities
Vector identities are useful tools in vector calculus that simplify complex vector operations. They are rules or properties that apply to vector arithmetic. One common identity used in vector calculus is the distributive law for cross products:
  • \(\mathbf{A} \times (\mathbf{B} + \mathbf{C}) = \mathbf{A} \times \mathbf{B} + \mathbf{A} \times \mathbf{C}\)
This identity highlights how a single vector can be crossed with the sum of two vectors, resulting in two separate cross products.
An example of using vector identities can be seen in problem (d) from the exercise, where we verify the equation
  • \((\mathbf{A}-\mathbf{B}) \times (\mathbf{A}+\mathbf{B}) = 2(\mathbf{A} \times \mathbf{B})\)
By expanding the left-hand side using vector identities and verifying the right-hand side multiplication, we can determine the truth of the statement. Vector identities thus serve as key simplifiers in solving vector problems.
Cross Product
The cross product is a binary operation on two vectors in three-dimensional space, producing another vector that is perpendicular to the plane of the input vectors.
The magnitude of the cross product is given by the formula:
  • \(||\mathbf{A} \times \mathbf{B}|| = ||\mathbf{A}|| ||\mathbf{B}|| \sin(\theta)\)
where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\). The direction follows the right-hand rule.
For example, in part (a) of the exercise, we evaluate \(\mathbf{i} \times (\mathbf{i} \times \mathbf{j})\), which involves the cross products:
  • \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \)
  • \( \mathbf{i} \times \mathbf{k} = -\mathbf{j} \)
These calculations help determine that this statement is false since the result is non-zero. Understanding cross products is vital, especially when dealing with perpendicular vectors and finding vector areas.
Scalar Triple Product
The scalar triple product is a calculation involving three vectors, resulting in a scalar. It is defined as follows:
  • \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\)
This operation gives the volume of the parallelepiped formed by the vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\). One interesting property of the scalar triple product is its cyclical nature:
  • \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}) = \mathbf{C} \cdot (\mathbf{A} \times \mathbf{B})\)
This symmetry was illustrated in part (b) of the exercise where both sides give the same result:
  • \(\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k})\)
  • \((\mathbf{i} \times \mathbf{j}) \cdot \mathbf{k} \)
In these expressions, both sides were found to be equal, confirming the statement. Mastery of the scalar triple product is crucial in calculations involving volumes and cyclic vector operations.
Vector Notation
Vector notation is a concise way to represent vectors and their operations in mathematics, especially in physics and engineering problems. A vector is typically denoted by a lowercase bold letter, such as \(\mathbf{a}\) or \(\mathbf{v}\), or using an arrow
  • \( \vec{a} \)
These notations signal that the quantity has both magnitude and direction.
For use in calculations, vectors can also be expressed in terms of unit vectors, commonly \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), which denote the standard basis in three-dimensional space:
  • \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \)
Using unit vectors makes it easy to perform operations such as the dot product or cross product.
For example, the problem set uses \(\mathbf{i}, \mathbf{j}, \mathbf{k} \) to represent orthogonal unit vectors, which simplifies verifying vector equations. Understanding vector notation is essential for resolving complex vector problems efficiently.

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Most popular questions from this chapter

Describe all points \((x, y)\) such that \(v=x \mathbf{i}+y\) satisfies (a) \(|\mathbf{v}|=2\) (b) \(|v-i|=2\) (c) \(\mathbf{v} \cdot \mathbf{i}=2\) (d) \(\mathbf{v} \cdot \mathbf{i}=|\mathbf{v}|\)

Use the matrices \(A, B, C\). $$A=\left[\begin{array}{lll}1 & 4 & 0 \\ 0 & 2 & 6 \\ 0 & 0 & 3\end{array}\right] \quad B=\left[\begin{array}{lll}0 & 0 & 1 \\ 2 & 1 & 0 \\\ 6 & 4 & 0\end{array}\right] \quad C=\left[\begin{array}{rrr}1 & -1 & -3 \\\ -1 & 2 & 0 \\ 0 & -1 & 3\end{array}\right]$$ Choose any right side \(\mathrm{d}\) that is not perpendicular to \(v=\) (1,1,1) . Show by elimination (reach an impossible equation) that \(C \mathbf{u}=\mathbf{d}\) has no solution.

(a) The triple cross product \((A \times B) \times C\) is in the plane of \(\mathbf{A}\) and \(\mathbf{B}\), because it is perpendicular to the cross product _____. (b) Compute \((\mathbf{A} \times \mathbf{B}) \times \mathbf{C}\) when \(\mathbf{A}=a_{2} \mathbf{i}+a_{2} \mathbf{j}+a_{3} \mathbf{k}, \mathbf{B}=\) \(b_{1} \mathbf{i}+b_{2} \mathbf{j}+b_{3} \mathbf{k}, \mathbf{C}=\mathbf{i}\) (c) Compute \((\mathbf{A} \cdot \mathbf{C}) \mathbf{B}-(\mathbf{B} \cdot \mathbf{C}) \mathbf{A}\) when \(\mathbf{C}=\mathbf{i}\). The answers in (b) and (c) should agree. This is also true if \(\mathbf{C}=\mathbf{j}\) or \(\mathbf{C}=\) \(\mathbf{k}\) or \(\mathbf{C}=c_{1} \mathbf{i}+c_{2} \mathbf{j}+c_{3} \mathbf{k} .\) That proves the tricky formula $$(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}=(\mathbf{A} \cdot \mathbf{C}) \mathbf{B}-(\mathbf{B} \cdot \mathbf{C}) \mathbf{A}$$

Suppose \(u\) is not the zero vector but \(A \mathbf{u}=0\). Then \(A^{-1}\) can't exist: It would multiply ____ and produce \(\mathbf{u}\).

(a) If two planes are perpendicular (front wall and side wall), is every line in one plane perpendicular to every line in the other? (b) If a third plane is perpendicular to the first, it might be (parallel) (perpendicular) (at a \(45^{\circ}\) angle) to the second.

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