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Chapter 12: Problem 70

Determine whether the following statements are true using a proof or counterexample. Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$\mathbf{u} \times(\mathbf{u} \times \mathbf{v})=\mathbf{0}$$

Short Answer

Expert verified
Answer: No, the statement is true if and only if \(\mathbf{u}\) is orthogonal to \(\mathbf{v}\).

Step by step solution

01

Refresh the properties of cross product

The cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) in \(\mathbb{R}^3\) is given by: $$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix}$$ Some important properties of the cross product are: 1. \(\mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a})\) 2. \(\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\) 3. \((\lambda \mathbf{a}) \times \mathbf{b} = \lambda (\mathbf{a} \times \mathbf{b})\) 4. \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\) if and only if \(\mathbf{a}\) and \(\mathbf{b}\) are parallel.
02

Expand the given expression

We are given the expression \(\mathbf{u} \times (\mathbf{u} \times \mathbf{v})\). First, we need to find the cross product of \(\mathbf{u} \times \mathbf{v}\): $$\mathbf{u} \times \mathbf{v} = \begin{pmatrix} u_2v_3 - u_3v_2 \\ u_3v_1 - u_1v_3 \\ u_1v_2 - u_2v_1 \end{pmatrix}$$ Now, we need to find the cross product of \(\mathbf{u}\) and the vector above: $$\mathbf{u} \times (\mathbf{u} \times \mathbf{v}) = \begin{pmatrix} u_2(u_3v_1 - u_1v_3) - u_3(u_1v_2 - u_2v_1) \\ u_3(u_1v_2 - u_2v_1) - u_1(u_2v_3 - u_3v_2) \\ u_1(u_2v_3 - u_3v_2) - u_2(u_3v_1 - u_1v_3) \end{pmatrix}$$
03

Check if the given expression results in the zero vector

Now, we need to determine if the result of the previous step is the zero vector. The elements of the resulting vector are: $$\begin{aligned} &u_2(u_3v_1 - u_1v_3) - u_3(u_1v_2 - u_2v_1) \\ &u_3(u_1v_2 - u_2v_1) - u_1(u_2v_3 - u_3v_2) \\ &u_1(u_2v_3 - u_3v_2) - u_2(u_3v_1 - u_1v_3) \end{aligned}$$ After some simplifications, we get: $$\begin{aligned} &u_1(u_1v_1) + u_2(u_2v_2) + u_3(u_3v_3) \\ &u_1(u_1v_1) + u_2(u_2v_2) + u_3(u_3v_3) \\ &u_1(u_1v_1) + u_2(u_2v_2) + u_3(u_3v_3) \end{aligned}$$ Notice that each component is the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), i.e., \(\mathbf{u} \cdot \mathbf{v}\). Since the dot product of two nonzero vectors is always zero if and only if the vectors are orthogonal, we can conclude that the statement \(\mathbf{u} \times (\mathbf{u} \times \mathbf{v})=\mathbf{0}\) is true if and only if \(\mathbf{u}\) is orthogonal to \(\mathbf{v}\).

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Most popular questions from this chapter

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