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Chapter 11: Problem 19

Compute the following cross products. Then make a sketch showing the two vectors and their cross product. $$-2 \mathbf{i} \times 3 \mathbf{k}$$

Short Answer

Expert verified
Question: Compute the cross product of the vectors \(\mathbf{A} = -2\mathbf{i}\) and \(\mathbf{B} = 3\mathbf{k}\), and sketch the two vectors along with their cross product. Answer: The cross product of the given vectors is \(\mathbf{A} \times \mathbf{B} = 6\mathbf{j}\). When sketching, vector \(\mathbf{A}\) goes 2 units in the negative X direction, vector \(\mathbf{B}\) goes 3 units in the positive Z direction, and their cross product, vector \(\mathbf{A} \times \mathbf{B}\), goes 6 units in the positive Y direction, being orthogonal to both \(\mathbf{A}\) and \(\mathbf{B}\).

Step by step solution

01

Calculate the cross product

The given vectors are: $$ \mathbf{A} = -2\mathbf{i} \quad \text{and} \quad \mathbf{B} = 3\mathbf{k} $$ To compute the cross product of these two 3-dimensional base vectors, we can express them as triplets and then use the determinant formula for cross products: $$ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 0 \\ 0 & 0 & 3 \end{vmatrix} $$ Expanding the determinant, we get: $$ \mathbf{A} \times \mathbf{B} = (0\cdot0 - 0\cdot3)\mathbf{i} - (0\cdot(-2) - 3(-2)\cdot0)\mathbf{j} + (0\cdot0 - 0\cdot0)\mathbf{k} = 0\mathbf{i} + 6\mathbf{j} + 0\mathbf{k} $$ Therefore, the cross product of the two given vectors is: $$ \mathbf{A} \times \mathbf{B} = 6\mathbf{j} $$
02

Sketch the given vectors and their cross product

To sketch the given vectors and their cross product, follow the steps below: 1. Start by drawing the 3 axes (X, Y, and Z). 2. Draw vector \(\mathbf{A} = -2\mathbf{i}\): From the origin, move 2 units in the negative X direction. 3. Draw vector \(\mathbf{B} = 3\mathbf{k}\): From the origin, go 3 units in the positive Z direction. 4. Draw vector \(\mathbf{A} \times \mathbf{B} = 6\mathbf{j}\): From the origin, move 6 units in the positive Y direction. The sketch should clearly show that the cross product \(\mathbf{A} \times \mathbf{B}\) is orthogonal to both \(\mathbf{A}\) and \(\mathbf{B}\), and that it lies in the Y direction.

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