/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Compute the following cross prod... [FREE SOLUTION] | 91影视

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Chapter 12: 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
Answer: The cross product of the vectors $$-2\mathbf{i}$$ and $$3\mathbf{k}$$ is the 0 vector ($$0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}$$). This result indicates that the given vectors are parallel to the xz-plane and do not form a well-defined plane in 3D space, as the cross product does not have magnitude or direction.

Step by step solution

01

Calculate The Cross Product

To find the cross product of two vectors, we can use the formula: $$\mathbf{A} \times \mathbf{B} = (A_yB_z - A_zB_y) \mathbf{i} - (A_xB_z - A_zB_x) \mathbf{j} + (A_xB_y - A_yB_x) \mathbf{k}$$ Now, we have \(\mathbf{A} = -2\mathbf{i}\) and \(\mathbf{B} = 3\mathbf{k}\). We can calculate the cross product as follows: $$\begin{aligned} (-2\mathbf{i}) \times (3\mathbf{k}) &= (0 - 0) \mathbf{i} - ((-2)(0) - 0) \mathbf{j} + ((-2)(0) - 0) \mathbf{k} \\ &= 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} \end{aligned}$$ The cross product is thus: $$0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}$$, which is the 0 vector.
02

Sketch the Vectors and Their Cross Product

Now, we will sketch the two given vectors and their cross product: 1. Vector $$-2\mathbf{i}$$ is a horizontal vector pointing 2 units to the left (in the negative x-direction). 2. Vector $$3\mathbf{k}$$ is a vertical vector pointing 3 units upwards (in the positive z-direction). 3. The cross product $$0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}$$ is a point, or the origin (0, 0, 0), since it has no magnitude or direction. We can see that the two given vectors $$-2\mathbf{i}$$ and $$3\mathbf{k}$$ are parallel to the xz-plane and are not coplanar with any other vector in 3D space. The cross product being the 0 vector shows that the given vectors do not form a well-defined plane in 3D space.

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