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Chapter 12: Problem 20

Compute the following cross products. Then make a sketch showing the two vectors and their cross product. $$2 \mathbf{j} \times(-5) \mathbf{i}$$

Short Answer

Expert verified
Answer: The cross product of the two given vectors 2j and -5i is 10k. In a sketch, vectors 2j and -5i should lie on the xy-plane with 2j as a vertical arrow pointing upwards (2 units) and -5i as a horizontal arrow pointing left (5 units). The cross product, 10k, should be represented as an arrow perpendicular to both 2j and -5i, pointing outwards (10 units) along the z-axis.

Step by step solution

01

Write the given vectors in component form

The given vectors are $$2 \mathbf{j} \times(-5) \mathbf{i}$$. We can write them in component form as: $$\mathbf{A} = 0\mathbf{i} + 2\mathbf{j} + 0\mathbf{k}$$ $$\mathbf{B} = -5\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}$$
02

Compute the cross product of the vectors

We can calculate the cross product of vectors $$\mathbf{A}$$ and $$\mathbf{B}$$ using the following formula: $$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \\ \end{vmatrix}$$ Calculating the cross product of $$\mathbf{A}$$ and $$\mathbf{B}$$, we get: $$\mathbf{A} \times \mathbf{B} = (2*0 - 0*(-5))\mathbf{i} - (0*0 - 0*(-5))\mathbf{j} + (0*0 - (-5)*2)\mathbf{k}$$
03

Simplify the resulting vector

Simplifying the cross product, we get: $$\mathbf{A} \times \mathbf{B} = 0\mathbf{i} - 0\mathbf{j} + 10\mathbf{k}$$ So, the cross product of the given vectors is: $$\mathbf{A} \times \mathbf{B} = 10\mathbf{k}$$
04

Create a sketch of the two vectors and their cross product

To create a sketch, follow these instructions: 1. Represent vector $$\mathbf{A}$$ (2j) as a vertical arrow on the y-axis pointing upwards with length 2 units. 2. Represent vector $$\mathbf{B}$$ (-5i) as a horizontal arrow on the x-axis pointing left with length 5 units. 3. Represent vector $$\mathbf{A} \times \mathbf{B}$$ (10k) as an arrow perpendicular to both $$\mathbf{A}$$ and $$\mathbf{B}$$, pointing outwards with length 10 units (z-axis). In the sketch, you should see vectors $$\mathbf{A}$$ and $$\mathbf{B}$$ lying on the xy-plane, and their cross product $$\mathbf{A} \times \mathbf{B}$$ pointing outwards along the z-axis, forming a right-handed coordinate system.

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