https://file.aiquickdraw.com/mathbot/file/1730258060828fbqyxzsv.jpg

The problem provides two vectors \( \mathbf{A} = 5\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \mathbf{B} = 2\hat{i} - 3\hat{j} + 5\hat{k} \), and asks for three things:

### (a) Dot product between \( \mathbf{A} \) and \( \mathbf{B} \) and the angle between them
The dot product formula is:
\[
\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z
\]

Here, \( A_x = 5 \), \( A_y = 3 \), \( A_z = -4 \), \( B_x = 2 \), \( B_y = -3 \), and \( B_z = 5 \). So,
\[
\mathbf{A} \cdot \mathbf{B} = (5)(2) + (3)(-3) + (-4)(5) = 10 - 9 - 20 = -19
\]

The magnitude of the vectors are:
\[
|\mathbf{A}| = \sqrt{5^2 + 3^2 + (-4)^2} = \sqrt{25 + 9 + 16} = \sqrt{50}
\]
\[
|\mathbf{B}| = \sqrt{2^2 + (-3)^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38}
\]

The angle \( \theta \) between the vectors can be found using the formula:
\[
\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}
\]
\[
\cos \theta = \frac{-19}{\sqrt{50} \cdot \sqrt{38}} = \frac{-19}{\sqrt{1900}} = \frac{-19}{43.58899}
\]
\[
\theta = \cos^{-1} \left( \frac{-19}{43.58899} \right) \approx 116.94^\circ
\]

### (b) Cross product between \( \mathbf{A} \) and \( \mathbf{B} \), and proving \( \mathbf{A} \times \mathbf{B} = - (\mathbf{B} \times \mathbf{A}) \)

The cross product formula in component form is:
\[
\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & -4 \\ 2 & -3 & 5 \end{vmatrix}
\]

This determinant gives:
\[
\mathbf{A} \times \mathbf{B} = \hat{i}(3 \times 5 - (-4) \times (-3)) - \hat{j}(5 \times 5 - (-4) \times 2) + \hat{k}(5 \times (-3) - 3 \times 2)
\]
\[
\mathbf{A} \times \mathbf{B} = \hat{i}(15 - 12) - \hat{j}(25 + 8) + \hat{k}(-15 - 6)
\]
\[
\mathbf{A} \times \mathbf{B} = 3\hat{i} - 33\hat{j} - 21\hat{k}
\]

Now for \( \mathbf{B} \times \mathbf{A} \):
\[
\mathbf{B} \times \mathbf{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 5 \\ 5 & 3 & -4 \end{vmatrix}
\]
\[
\mathbf{B} \times \mathbf{A} = \hat{i}((-3)(-4) - (5)(3)) - \hat{j}((2)(-4) - (5)(5)) + \hat{k}((2)(3) - (-3)(5))
\]
\[
\mathbf{B} \times \mathbf{A} = \hat{i}(12 - 15) - \hat{j}(-8 - 25) + \hat{k}(6 + 15)
\]
\[
\mathbf{B} \times \mathbf{A} = -3\hat{i} + 33\hat{j} + 21\hat{k}
\]

As expected:
\[
\mathbf{A} \times \mathbf{B} = - (\mathbf{B} \times \mathbf{A})
\]

### (c) Finding the value of \( \mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) \)

Since the cross product \( \mathbf{A} \times \mathbf{B} \) is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \), the dot product of \( \mathbf{B} \) with \( \mathbf{A} \times \mathbf{B} \) will always be zero. Therefore:
\[
\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = 0
\]

Would you like more details or explanations on any part of this solution?

Here are 5 related questions:
1. What is the physical meaning of the dot product in vector mechanics?
2. How can we interpret the cross product geometrically?
3. Can the dot product of two vectors ever be negative?
4. What conditions must be met for two vectors to be perpendicular based on their dot product?
5. In what cases will the cross product of two vectors result in a zero vector?

**Tip:** The cross product results in a vector that is perpendicular to both of the original vectors. This property is useful in determining planes and rotations in 3D space.

Given that A = 5i + 3j - 4k and B = 2i - 3j + 5k, (a) Find the dot product between A and B and the angle between them. (b) Find the cross product between A & B and prove that (A × B) = - (B × A). (c) Find the value of B ⋅ (A × B).

Learn how to calculate the dot product and cross product of two vectors A = 5i + 3j - 4k and B = 2i - 3j + 5k. This problem covers vector algebra concepts, including the dot product formula, cross product using determinants, and finding the angle between vectors. Also, verify the anti-commutative property A × B = - (B × A) and compute B ⋅ (A × B).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation