Question

\(\quad\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)=\ldots\) (A) 0 (B) \(A^{2}+B^{2}\) (C) \(\sqrt{\left(A^{2}+B^{2}\right)}\) (D) \(\mathrm{A}^{2} \mathrm{~B}^{2}\)

Short answer

The short answer based on the step-by-step solution is: The expression \[\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot \left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)\] is equal to 0 (choice (A)).

Step-by-step solution

Write out the given expression

We need to find the value of the given expression, which is: \[\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot \left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)\]

Recall properties of dot and cross product

Remember that the dot product has the following properties: 1. Distributive: \(\mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow})= \mathrm{A}^{\rightarrow}\cdot \mathrm{B}^{\rightarrow}+\mathrm{A}^{\rightarrow}\cdot \mathrm{C}^{\rightarrow}\) 2. Commutative: \(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow} = \mathrm{B}^{\rightarrow} \cdot \mathrm{A}^{\rightarrow}\) As for the cross product, we have the following properties: 1. Anticommutative: \(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow} = -\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow}\) 2. Distributive: \(\mathrm{A}^{\rightarrow} \times (\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow}) = \mathrm{A}^{\rightarrow}\times \mathrm{B}^{\rightarrow} + \mathrm{A}^{\rightarrow}\times \mathrm{C}^{\rightarrow}\)

Apply the dot product distributive property

Applying the dot product distributive property, the given expression becomes: \[(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow})\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow}) = \mathrm{A}^{\rightarrow}\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow}) + \mathrm{B}^{\rightarrow}\cdot(\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow})\]

Apply the anticommutative property of the cross product

For both terms on the right-hand side of the equation, the dot product of a cross product is involved, so we can write: \[\mathrm{A}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}) = - \mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow})\] \[\mathrm{B}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}) = -\mathrm{B}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times \mathrm{A}^{\rightarrow})\] Now, the expression from Step 3 becomes: \[-(\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))+(-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))\]

Identify the scalar triple product and its property

The scalar triple product is defined as: \[\mathrm{A}^{\rightarrow} \cdot (\mathrm{B}^{\rightarrow} \times\mathrm{C}^{\rightarrow}) = \mathrm{B}^{\rightarrow} \cdot (\mathrm{C}^{\rightarrow} \times\mathrm{A}^{\rightarrow}) = \mathrm{C}^{\rightarrow} \cdot (\mathrm{A}^{\rightarrow}\times\mathrm{B}^{\rightarrow})\] Now we see that the terms in the expression from Step 4 are scalar triple products: \[-(\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))=-\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})\] \[-(\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow}))=-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})\] Since each term on the right-hand side is equal, their sum will be zero: \[-\mathrm{A}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})-\mathrm{B}^{\rightarrow}\cdot(\mathrm{B}^{\rightarrow}\times\mathrm{A}^{\rightarrow})=0\] Therefore, the original expression is equal to 0, which corresponds to choice (A).

Key concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It combines two vectors and yields a scalar result. For two vectors, \( \mathbf{A} \) and \( \mathbf{B} \), the dot product is calculated as:

• \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \)

where \( \theta \) is the angle between the vectors.
This product measures how much of one vector goes in the direction of another. If the angle is 90 degrees, the dot product is zero. This is because the vectors are perpendicular and do not overlap in any direction.
The dot product is also distributive and commutative. For example, if there's a vector \( \mathbf{C} \),:

• \( \mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C} \)
• \( \mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A} \)

This makes the dot product a versatile tool in vector calculations and useful in various applications like physics and engineering.

Cross Product

The cross product, another essential operation in vector algebra, produces a vector that is perpendicular to the plane containing the two vectors. Given vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cross product is calculated as:

• \( \mathbf{A} \times \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \sin(\theta) \ \mathbf{n} \)

Here, \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{B} \) and \( \mathbf{n} \) is a unit vector perpendicular to the plane created by \( \mathbf{A} \) and \( \mathbf{B} \).
Importantly, the cross product is not commutative, meaning \( \mathbf{A} \times \mathbf{B} eq \mathbf{B} \times \mathbf{A} \). Instead, \( \mathbf{A} \times \mathbf{B} = -\mathbf{B} \times \mathbf{A} \), reflecting the anticommutative property.
The magnitude of the cross product is the area of the parallelogram that the vectors span. This makes it useful for determining orthogonal directions, essential in fields like computer graphics and physics.

Vector Algebra

Vector algebra involves operations that encompass vectors, their magnitudes, and directions. Vectors in mathematics are quantities that have both direction and magnitude. The study of vector algebra includes understanding basic operations like addition, subtraction, dot products, and cross products.
Vector addition is straightforward, combining two vectors' corresponding components:

• If \( \mathbf{A} = (a_1, a_2, a_3) \) and \( \mathbf{B} = (b_1, b_2, b_3) \), the sum is \( \mathbf{A} + \mathbf{B} = (a_1 + b_1, a_2 + b_2, a_3 + b_3) \).

Subtraction follows a similar rule. Dot and cross products build on these concepts to offer more complex calculations, allowing for determining angles between vectors and orthogonal projections.Other principles include the distributive property, helping expand vector expressions further, such as \( \mathbf{A} \times (\mathbf{B} + \mathbf{C}) = \mathbf{A} \times \mathbf{B} + \mathbf{A} \times \mathbf{C} \). This enhances the manipulation and solving of vector equations, a powerful tool in real-world applications from navigation to physics.

Anticommutative Property

The anticommutative property is a key feature of the cross product in vector algebra. Unlike the dot product, the cross product exhibits this property. For two vectors, \( \mathbf{A} \) and \( \mathbf{B} \), the anticommutative property is expressed as:

• \( \mathbf{A} \times \mathbf{B} = - (\mathbf{B} \times \mathbf{A}) \)

This means reversing the order of the vectors in a cross product inverts the direction of the resulting vector. Essentially, changing the sequence yields a vector that points in the opposite direction.
This characteristic is crucial in ensuring that the cross product results in perpendicular vector determination, aligning with the right-hand rule. This rule helps define the direction of the vector when computing cross products and is used extensively in physics, particularly in electromagnetism and rotational dynamics.
Understanding this property enhances comprehension of vector behavior in three-dimensional space, a crucial aspect in many scientific and engineering disciplines.