Question

Given vector A = 31+4j2k and B = 21-6j+3k, find

a. AxB

b. The angle & between A and B​

Answer 1

a. The cross product
`\( \mathbf{A} * \mathbf{B} = 26\mathbf{i} - 5\mathbf{j} - 18\mathbf{k} \).`

b. The angle
`\( \theta \)` between vectors A and B is calculated using the dot product and magnitudes, yielding
`\[ \theta = \arccos\left((-12)/(√(29) \cdot 7)\right) \]`.

**a. A × B (Cross Product):**

The cross product of two vectors A and B in three-dimensional space is determined by the formula:

`\[ A * B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \]`

Given
`\( \mathbf{A} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \) and \( \mathbf{B} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k} \)`, we can compute the cross product:

`\[ \mathbf{A} * \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 2 \\ 2 & -6 & 3 \end{vmatrix} \]`

Expanding the determinant:

`\[ \mathbf{A} * \mathbf{B} = \mathbf{i}(4 * 3 - (-6) * 2) - \mathbf{j}(3 * 3 - 2 * 2) + \mathbf{k}(3 * (-6) - 4 * 2) \]`

`\[ \mathbf{A} * \mathbf{B} = 26\mathbf{i} - 5\mathbf{j} - 18\mathbf{k} \]`

**b. Angle
`\( \theta \)` between A and B:**

The dot product
`(\( \cdot \))` and magnitudes ( |A|, |B| ) are related to the angle
`\( \theta \)` between two vectors A and B through:

`\[ \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}} \cdot \]`

Given
`\( \mathbf{A} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \) and \( \mathbf{B} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k} \)`, we calculate:

`\[ \mathbf{A} \cdot \mathbf{B} = (3 * 2) + (4 * (-6)) + (2 * 3) \]`

`\[ \mathbf{A} \cdot \mathbf{B} = 6 - 24 + 6 = -12 \]`

`\[ |A| = √(3^2 + 4^2 + 2^2) = √(9 + 16 + 4) = √(29) \]`

`\[ |B| = √(2^2 + (-6)^2 + 3^2) = √(4 + 36 + 9) = √(49) = 7 \]`

`\[ \cos(\theta) = (-12)/(√(29) \cdot 7) \]`

`\[ \theta = \arccos\left((-12)/(√(29) \cdot 7)\right) \]`