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Chapter 12: Problem 9

Sketch the following vectors \(\mathbf{u}\) and \(\mathbf{v} .\) Then compute \(|\mathbf{u} \times \mathbf{v}|\) and show the cross product on your sketch. $$\mathbf{u}=\langle 0,-2,0\rangle, \mathbf{v}=\langle 0,1,0\rangle$$

Short Answer

Expert verified
Question: Compute the cross product and its magnitude of the given two vectors: \(\mathbf{u}=\langle 0,-2,0\rangle\) and \(\mathbf{v}=\langle 0,1,0\rangle\), and represent them in a sketch. Answer: The cross product of \(\mathbf{u}\) and \(\mathbf{v}\) is \(\mathbf{u} \times \mathbf{v} = \langle 0, 0, 0 \rangle\). The magnitude of this cross product is |\(\mathbf{u} \times \mathbf{v}\)| = \(0\). In the sketch, both vectors \(\mathbf{u}\) and \(\mathbf{v}\) are vertical and parallel to the y-axis with the tail at the origin, and the cross product vector is a zero vector which lies at the origin.

Step by step solution

01

Sketch the Vectors

To sketch the given vectors \(\mathbf{u}=\langle 0,-2,0\rangle\) and \(\mathbf{v}=\langle 0,1,0\rangle\), we observe that both vectors only have components in the y-axis. Thus, they are vertical and parallel to the y-axis. In the xy-plane, place the tail of both vectors at the origin and draw arrows downward (for \(\mathbf{u}\)) and upward (for \(\mathbf{v}\)).
02

Compute the Cross Product

To compute the cross product, use the formula: $$\mathbf{u} \times \mathbf{v} = \langle (u_2v_3 - u_3v_2), (u_3v_1 - u_1v_3), (u_1v_2 - u_2v_1) \rangle$$ Substitute the given values to get: $$\mathbf{u} \times \mathbf{v} = \langle (-2\cdot0 - 0\cdot1), (0\cdot0 - 0\cdot0), (0\cdot1 - (-2)\cdot0) \rangle$$ $$\mathbf{u} \times \mathbf{v} = \langle 0, 0, 0 \rangle$$
03

Compute the Magnitude of the Cross Product

Since the cross product is \(\langle 0, 0, 0 \rangle\), its magnitude is given by: $$|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + 0^2} = 0$$
04

Include the Cross Product Vector in the Sketch

The cross product vector is a zero vector. This means that it has no direction and its magnitude is zero. Hence, it lies at the origin. In the sketch, simply label the origin as \(\mathbf{u} \times \mathbf{v}\).

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Most popular questions from this chapter

An object moves clockwise around a circle centered at the origin with radius \(5 \mathrm{m}\) beginning at the point (0,5) a. Find a position function \(\mathbf{r}\) that describes the motion if the object moves with a constant speed, completing 1 lap every 12 s. b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{-t}\)

A 500-kg load hangs from three cables of equal length that are anchored at the points \((-2,0,0),(1, \sqrt{3}, 0),\) and \((1,-\sqrt{3}, 0) .\) The load is located at \((0,0,-2 \sqrt{3}) .\) Find the vectors describing the forces on the cables due to the load.

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).

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. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).

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 \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)
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