Compute the Cross Product of Vectors Using Determinants

Math Problem Statement

\mathbf{B} = \mathbf{T} \times \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 0 & 16 \\ 0 & -16 & 0 \end{vmatrix} = (0 \cdot 0 - 16 \cdot (-16)) \mathbf{i} - (-4 \cdot 0 - 16 \cdot 0) \mathbf{j} + (-4 \cdot (-16) - 0 \cdot 0) \mathbf{k}

Let's compute the cross product step by step:

The cross product $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ is given by the determinant:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 0 & 16 \\ 0 & -16 & 0 \end{vmatrix}$$ Expanding this determinant: $$\mathbf{B} = \mathbf{i} \left( 0 \cdot 0 - 16 \cdot (-16) \right) - \mathbf{j} \left( -4 \cdot 0 - 16 \cdot 0 \right) + \mathbf{k} \left( -4 \cdot (-16) - 0 \cdot 0 \right)$$ Simplifying each component: For the $$\mathbf{i}$$ component: $$\mathbf{i} (0 - (-256)) = \mathbf{i} \cdot 256$$ For the $$\mathbf{j}$$ component: $$- \mathbf{j} (0 - 0) = 0$$ For the $$\mathbf{k}$$ component: $$\mathbf{k} (64 - 0) = \mathbf{k} \cdot 64$$ Therefore, the cross product is: $$\mathbf{B} = 256\mathbf{i} + 0\mathbf{j} + 64\mathbf{k}$$ So, $$\mathbf{B} = 256\mathbf{i} + 64\mathbf{k}$$ Would you like any further details or have any questions about this computation? Here are 5 questions you might consider next: 1. How do you find the magnitude of a vector? 2. What are some applications of the cross product in physics? 3. How do you determine if two vectors are perpendicular? 4. How can you express a vector in polar coordinates? 5. What is the dot product and how is it different from the cross product? **Tip:** When computing determinants, always remember to carefully handle signs and zeros to avoid common errors.

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Vectors
Cross Product
Determinants

Advanced High School

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