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Chapter 12: Problem 11

Sketch the following vectors \(\mathbf{u}\) and \(\mathbf{v} .\) Then compute \(|\mathbf{u} \times \mathbf{v}|\) and show the cross product on your sketch. $$\mathbf{u}=\langle 3,3,0\rangle, \mathbf{v}=\langle 3,3,3 \sqrt{2}\rangle$$

Short Answer

Expert verified
In this exercise, we first sketched the vectors \(\mathbf{u}\) and \(\mathbf{v}\) in three-dimensional space. Then we computed their cross product \(\mathbf{w} = \mathbf{u}\times \mathbf{v} = (0, -9\sqrt{2}, 0)\). The magnitude of the cross product is \(|\mathbf{w}| = 9\sqrt{2}\). Finally, we sketched the cross product vector \(\mathbf{w}\) on the same plot as the original vectors, showing it to be perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).

Step by step solution

01

Sketch vectors \(\mathbf{u}\) and \(\mathbf{v}\)

To sketch the vectors, plot them as arrows in three-dimensional space. Vector \(\mathbf{u}\) goes from the origin to the point \((3, 3, 0)\), while vector \(\mathbf{v}\) goes from the origin to the point \((3, 3, 3\sqrt{2})\). Both vectors lie in the plane formed by the \(x,y\) and \(z\) axes.
02

Compute the cross product \(\mathbf{u} \times \mathbf{v}\)

To compute the cross product, use the formula: $$\begin{pmatrix} i & j & k \\3 & 3 & 0 \\3 & 3 & 3\sqrt{2}\end{pmatrix}$$ Calculating the determinant, we obtain: $$\mathbf{w} = \mathbf{u}\times \mathbf{v}= (-9\sqrt{2}k)-(0)-(0)=(0,-9\sqrt{2},0)$$
03

Find the magnitude of the cross product \(|\mathbf{u} \times \mathbf{v}|\)

To find the magnitude, use the formula \(|\mathbf{w}|=\sqrt{(0)^2+ (-9\sqrt{2})^2 +(0)^2} = 9\sqrt{2}\).
04

Sketch the cross product vector

The cross product vector \(\mathbf{w}=(0,-9\sqrt{2},0)\) goes from the origin to the point \((0, -9\sqrt{2}, 0)\). This vector is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\), and it lies in the \(y\)-axis direction. Sketch the cross product vector on the same plot as the original vectors.

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