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Find all vectors \(\mathbf{u}\) that satisfy the equation $$ \langle 1,1,1\rangle \times \mathbf{u}=\langle 0,0,1\rangle $$

Short Answer

Expert verified
Answer: No, there are no vectors 𝐮 that satisfy the given equation.

Step by step solution

01

Understand the cross product formula

The cross product of two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\) is defined as the vector \(\mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\). The cross product is orthogonal to both original vectors and has a magnitude equal to the area of the parallelogram spanned by \(\mathbf{a}\) and \(\mathbf{b}\).
02

Set up the equation

Let's set up the equation using the cross product formula: $$ \langle 1,1,1\rangle \times \mathbf{u} = \langle 0, 0, 1\rangle $$ Let \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\), and apply the cross product formula to get the following equation: $$ \langle 1u_3 - 1u_2, 1u_1 - 1u_3, 1u_2 - 1u_1 \rangle = \langle 0, 0, 1\rangle $$
03

Solve for u components

In order for the cross product to equal \(\langle 0,0,1\rangle\), the corresponding components must be equal. Thus, we have the following system of equations: (1) \(1u_3 - 1u_2 = 0\)\ (2) \(1u_1 - 1u_3 = 0\)\ (3) \(1u_2 - 1u_1 = 1\) First, let's solve equation (1) for \(u_3\), and equation (3) for \(u_2\): \(u_3 = u_2\)\ \(u_2 = u_1 + 1\) Now substitute \(u_3\) with \(u_2\) in equation (2): \(u_1 - u_2 = 0\) And finally, substitute \(u_2 = u_1 + 1\) into the above equation: \(u_1 - (u_1 + 1) = 0\) This simplifies to: \(-1 = 0\) Thus, we see that there is no solution for the components of \(\mathbf{u}\) that will satisfy all three equations. Therefore, there are no vectors \(\mathbf{u}\) that satisfy the given equation \(\langle 1,1,1\rangle \times \mathbf{u} = \langle 0, 0, 1\rangle\).

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The line \(\mathbf{r}(t)=\langle 3,-1,4\rangle+t\langle 6,-2,8\rangle\) passes through the origin. b. Any two nonparallel lines in \(\mathbb{R}^{3}\) intersect. c. The curve \(\mathbf{r}(t)=\left\langle e^{-t}, \sin t,-\cos t\right\rangle\) approaches a circle as \(t \rightarrow \infty\). d. If \(\mathbf{r}(t)=e^{-t^{2}}\langle 1,1,1\rangle\) then \(\lim _{t \rightarrow \infty} \mathbf{r}(t)=\lim _{t \rightarrow-\infty} \mathbf{r}(t)\).

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