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Chapter 12: Problem 13

Compute \(|\mathbf{u} \times \mathbf{v}|\) if \(\mathbf{u}\) and \(\mathbf{v}\) are unit vectors and the angle between \(\mathbf{u}\) and \(\mathbf{v}\) is \(\pi / 4\)

Short Answer

Expert verified
Answer: \(\frac{\sqrt{2}}{2}\)

Step by step solution

01

Recall the formula for the magnitude of the cross product

We will use the following formula to calculate the magnitude of the cross product of the given unit vectors \(\mathbf{u}\) and \(\mathbf{v}\): $$|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin{\theta}$$ Here, \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), and \(\theta\) is the angle between them.
02

Substitute the given values into the formula

As \(\mathbf{u}\) and \(\mathbf{v}\) are unit vectors, their magnitudes are both equal to 1. Also, we are given that the angle between them \(\theta\) is \(\pi/4\). We can plug these values directly into the formula: $$|\mathbf{u} \times \mathbf{v}| = (1)(1)\sin{\left(\frac{\pi}{4}\right)}$$
03

Calculate the magnitude of the cross product

Now, we just need to evaluate the sine function and perform the multiplication: $$|\mathbf{u} \times \mathbf{v}| = \sin{\left(\frac{\pi}{4}\right)}$$ The sine function of \(\pi/4\) is equal to \(\frac{\sqrt{2}}{2}\), so we have: $$|\mathbf{u} \times \mathbf{v}| = \frac{\sqrt{2}}{2}$$ Hence, the magnitude of the cross product of the given unit vectors is \(\frac{\sqrt{2}}{2}\).

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Most popular questions from this chapter

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)

Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

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. $$\begin{aligned} \mathbf{r}(t)=&(2 \cos t+2 \sin t) \mathbf{i}+(-\cos t+2 \sin t) \mathbf{j} \\\ &+(\cos t-2 \sin t) \mathbf{k} \end{aligned}$$

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. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).
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