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Chapter 12: Problem 17

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

Short Answer

Expert verified
Answer: The cross product of the vectors -j and k is the vector i. Geometrically, this vector is orthogonal to the plane formed by the two original vectors, -j and k.

Step by step solution

01

Write the given vectors in component form

Let's write the given vectors -j and k in their component form: $$-\mathbf{j} = \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix} \quad \text{and} \quad \mathbf{k} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$
02

Compute the cross product using the determinant formula

To compute the cross product, we use the determinant formula, which consists of the vector components in a 3x3 matrix along with the unit vectors i, j, and k. The cross product looks like this: $$-\mathbf{j} \times \mathbf{k} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$ To compute the determinant, we take the sum of three values, each of which is the product of an element from the first row, its corresponding cofactor, and the unit vectors: $$(-1) \cdot \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} \mathbf{i} - (1) \cdot \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} \mathbf{j} + (0) \cdot \begin{vmatrix} 0 & -1 \\ 0 & 0 \end{vmatrix} \mathbf{k}$$
03

Compute the determinant and simplify the expression

We now compute the determinants and simplify the expression: $$(-1)(-1\cdot1 - 0\cdot0) \mathbf{i} - (1)(0\cdot1 - 0\cdot0) \mathbf{j} + (0)(0\cdot0 - 0\cdot(-1)) \mathbf{k}$$ $$= 1 \mathbf{i} - 0 \mathbf{j} + 0 \mathbf{k}$$ So, the cross product -j 脳 k is given by: $$-\mathbf{j} \times \mathbf{k} = \mathbf{i}$$
04

Sketch the vectors and their cross product

To make a sketch showing the two vectors and their cross product, follow these steps: 1. Draw the coordinate axes (x, y, z) on a sheet of paper or using a graphics tool. 2. Place the origin (0,0,0) at the intersection of the axes. 3. Draw the first vector, -j, starting from the origin and going down (-1 unit) along the y-axis. 4. Draw the second vector, k, starting from the origin and going up (1 unit) along the z-axis. 5. Draw the cross product, i, which is the result vector starting from the origin and going to the right (1 unit) along the x-axis. The sketch should show that the new cross product vector, i, is orthogonal to the plane formed by the two original vectors, -j and k.

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Most popular questions from this chapter

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{array}{l} \mathbf{r}(t)=\langle 4+t,-2 t, 1+3 t\rangle ;\\\ \mathbf{R}(s)=\langle 1-7 s, 6+14 s, 4-21 s\rangle \end{array}$$

Find the point (if it exists) at which the following planes and lines intersect. $$z=4 ; \mathbf{r}(t)=\langle 2 t+1,-t+4, t-6\rangle$$

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Consider an object moving along the circular trajectory \(\mathbf{r}(t)=\langle A \cos \omega t, A \sin \omega t\rangle,\) where \(A\) and \(\omega\) are constants. a. Over what time interval \([0, T]\) does the object traverse the circle once? b. Find the velocity and speed of the object. Is the velocity constant in either direction or magnitude? Is the speed constant? c. Find the acceleration of the object. d. How are the position and velocity related? How are the position and acceleration related? e. Sketch the position, velocity, and acceleration vectors at four different points on the trajectory with \(A=\omega=1\)
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