https://file.aiquickdraw.com/mathbot/file/1729485805815qvqkfb3u.jpg

Let's break down the problem step by step.

### Given:
\[
\mathbf{A} = 5\hat{i} + 3\hat{j} - 4\hat{k}
\]
\[
\mathbf{B} = 2\hat{i} - 3\hat{j} + 5\hat{k}
\]

### (a) Find the Dot product between **A** and **B**.

The dot product formula is:
\[
\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z
\]

Substitute the components:
\[
\mathbf{A} \cdot \mathbf{B} = (5)(2) + (3)(-3) + (-4)(5)
\]
\[
= 10 - 9 - 20
\]
\[
\mathbf{A} \cdot \mathbf{B} = -19
\]

### (b) Find the cross product between **A** and **B** and prove that \(\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})\).

The cross product formula for vectors \(\mathbf{A}\) and \(\mathbf{B}\) is:
\[
\mathbf{A} \times \mathbf{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{vmatrix}
\]

Substitute the values of \(\mathbf{A}\) and \(\mathbf{B}\):
\[
\mathbf{A} \times \mathbf{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 & 3 & -4 \\
2 & -3 & 5
\end{vmatrix}
\]
Now expand the determinant:
\[
= \hat{i} \left( 3(5) - (-4)(-3) \right) - \hat{j} \left( 5(5) - (-4)(2) \right) + \hat{k} \left( 5(-3) - 3(2) \right)
\]
\[
= \hat{i} \left( 15 - 12 \right) - \hat{j} \left( 25 - (-8) \right) + \hat{k} \left( -15 - 6 \right)
\]
\[
= \hat{i}(3) - \hat{j}(33) + \hat{k}(-21)
\]
\[
\mathbf{A} \times \mathbf{B} = 3\hat{i} - 33\hat{j} - 21\hat{k}
\]

To prove \(\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})\), we calculate \(\mathbf{B} \times \mathbf{A}\):
\[
\mathbf{B} \times \mathbf{A} = -\mathbf{A} \times \mathbf{B}
\]
Using the fact that the cross product is anti-commutative, we already know that \(\mathbf{B} \times \mathbf{A}\) should equal:
\[
\mathbf{B} \times \mathbf{A} = - (3\hat{i} - 33\hat{j} - 21\hat{k}) = -3\hat{i} + 33\hat{j} + 21\hat{k}
\]
Thus, the proof holds.

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

We know that the result of \(\mathbf{A} \times \mathbf{B} = 3\hat{i} - 33\hat{j} - 21\hat{k}\). To compute \(\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B})\), we use the dot product formula:
\[
\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = B_x(3) + B_y(-33) + B_z(-21)
\]
Substitute the components of \(\mathbf{B}\):
\[
= (2)(3) + (-3)(-33) + (5)(-21)
\]
\[
= 6 + 99 - 105
\]
\[
\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = 0
\]

Since the result is zero, it means \(\mathbf{B}\) is orthogonal to \(\mathbf{A} \times \mathbf{B}\).

---

Do you need further clarifications or details on any step?

Here are 5 related questions to explore this further:
1. How do you compute the magnitude of a vector?
2. What is the geometric interpretation of the dot product and cross product?
3. How can you find the angle between two vectors using the dot product?
4. What does it mean geometrically if the cross product of two vectors is zero?
5. Can you prove that the magnitude of \(\mathbf{A} \times \mathbf{B}\) gives the area of the parallelogram formed by \(\mathbf{A}\) and \(\mathbf{B}\)?

**Tip:** For quick verification of your results, use unit vectors and their properties, such as the anti-commutative nature of the cross product.

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 compute the dot product and cross product of vectors A = 5i + 3j - 4k and B = 2i - 3j + 5k. Explore key concepts such as the angle between vectors, the anti-commutative property of the cross product, and orthogonality. Suitable for college-level or advanced high school students.

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