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Chapter 12: Problem 4

If \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal, what is the magnitude of \(\mathbf{u} \times \mathbf{v} ?\)

Short Answer

Expert verified
Answer: When \(\textbf{u}\) and \(\textbf{v}\) are orthogonal, the magnitude of their cross product is the product of their magnitudes, which can be represented as \(|\textbf{u} \times \textbf{v}| = |\textbf{u}| \cdot |\textbf{v}|\).

Step by step solution

01

Understanding the Cross Product

Recall that the cross product of two vectors \(\textbf{u}\) and \(\textbf{v}\), denoted by \(\textbf{u} \times \textbf{v}\), is a new vector which is orthogonal to the plane formed by \(\textbf{u}\) and \(\textbf{v}\). The magnitude of the cross product is given by: \(|\textbf{u} \times \textbf{v}| = |\textbf{u}| \cdot |\textbf{v}| \cdot \sin{\theta}\) where \(|\textbf{u}|\) and \(|\textbf{v}|\) are the magnitudes of the vectors \(\textbf{u}\) and \(\textbf{v}\) respectively, and \(\theta\) is the angle between them.
02

Determine the Angle Between Orthogonal Vectors

If \(\textbf{u}\) and \(\textbf{v}\) are orthogonal, this means that the angle between them is \(90^{\circ}\) (or \(\frac{\pi}{2}\) radians). It is relevant because the sine of this angle is needed for the cross product magnitude calculation: \(\sin{90^{\circ}} = 1\)
03

Calculate the Magnitude of the Cross Product

Given that \(\textbf{u}\) and \(\textbf{v}\) are orthogonal, we can substitute their angle and the sine value in the formula for the magnitude of the cross product: \(|\textbf{u} \times \textbf{v}| = |\textbf{u}| \cdot |\textbf{v}| \cdot 1\) Thus: \(|\textbf{u} \times \textbf{v}| = |\textbf{u}| \cdot |\textbf{v}|\) So when \(\textbf{u}\) and \(\textbf{v}\) are orthogonal, the magnitude of their cross product is simply the product of their magnitudes.

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Most popular questions from this chapter

Trajectory with a sloped landing Assume an object is launched from the origin with an initial speed \(\left|\mathbf{v}_{0}\right|\) at an angle \(\alpha\) to the horizontal, where \(0 < \alpha < \frac{\pi}{2}\) a. Find the time of flight, range, and maximum height (relative to the launch point) of the trajectory if the ground slopes downward at a constant angle of \(\theta\) from the launch site, where \(0 < \theta < \frac{\pi}{2}\) b. Find the time of flight, range, and maximum height of the trajectory if the ground slopes upward at a constant angle of \(\theta\) from the launch site.

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned} \mathbf{r}(t)=&(2 \cos t+2 \sin t) \mathbf{i}+(-\cos t+2 \sin t) \mathbf{j} \\\ &+(\cos t-2 \sin t) \mathbf{k} \end{aligned}$$

The function \(f(x)=\sin n x,\) where \(n\) is a positive real number, has a local maximum at \(x=\pi /(2 n)\) Compute the curvature \(\kappa\) of \(f\) at this point. How does \(\kappa\) vary (if at all) as \(n\) varies?

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the Cauchy-Schwarz Inequality?
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