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Chapter 12: Problem 5

Explain how to use a determinant to compute \(\mathbf{u} \times \mathbf{v}\)

Short Answer

Expert verified
Question: Compute the cross product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) using determinants. Answer: To compute the cross product, form a \(3 \times 3\) matrix with the first row being the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), the second row being the components of \(\mathbf{u}\), and the third row being the components of \(\mathbf{v}\). Then, expand and compute the determinant to find the cross product: \(\mathbf{u} \times \mathbf{v} = (\begin{array}{l}u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1\end{array})\).

Step by step solution

01

Write the cross product determinant matrix

Let \(\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}\). To find the cross product using determinants, form a matrix where the first row are the unit vectors, \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), the second row is the components of \(\mathbf{u}\), and the third row is the components of \(\mathbf{v}\): $$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} $$
02

Expand and compute the determinant

Expand the determinant along the first row, which consists of the unit vectors. Since \(\mathbf{i}\) is in the first row and first column: $$ i \cdot \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}$$ Then, since \(\mathbf{j}\) is in the first row and second column: $$ j \cdot \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}$$ Finally, since \(\mathbf{k}\) is in the first row and third column: $$ k \cdot \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}$$ Now, compute each of the three determinants: $$ i\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} - j\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} + k\begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}$$
03

Evaluate the final cross product

Evaluate the resulting cross product of the two vectors \(\mathbf{u}\) and \(\mathbf{v}\): $$ \mathbf{u} \times \mathbf{v} = (\begin{array}{l}u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1\end{array}) $$ Now we have found the cross product of \(\mathbf{u}\) and \(\mathbf{v}\) using a determinant.

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Most popular questions from this chapter

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Given a fixed vector \(\mathbf{v},\) there is an infinite set of vectors \(\mathbf{u}\) with the same value of proj\(_{\mathbf{v}} \mathbf{u}\). Let \(\mathbf{v}=\langle 0,0,1\rangle .\) Give a description of all position vectors \(\mathbf{u}\) such that \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\operatorname{proj}_{\mathbf{v}}\langle 1,2,3\rangle\).

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 \(a b+c d+e f=0\).

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 \(\mathbf{v}\). \(\mathbf{u}=\langle-1,2,3\rangle, \mathbf{v}=\langle 2,1,1\rangle\)

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 \(\mathbf{v}\). \(\mathbf{u}=\langle-2,2\rangle, \mathbf{v}=\langle 2,1\rangle\)
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