Question

Find all vectors \(\mathbf{u}\) that satisfy the equation $$ \langle 1,1,1\rangle \times \mathbf{u}=\langle-1,-1,2\rangle $$

Short answer

Create a short answer based on the solution: To find a vector, \(\mathbf{u}\), such that its cross product with the given vector \(\langle 1,1,1\rangle\) equals the given vector \(\langle -1, -1, 2\rangle\), we first express the cross product of the two vectors and then set up a system of equations. After solving the system, we find that there are infinitely many solutions for \(\mathbf{u}\). Thus, we can express \(\mathbf{u}\) in terms of just one variable, say \(u_2\), as \(\mathbf{u} = \{\langle u_2 - 2, u_2, u_2 - 1 \rangle \ | \ u_2 \in \mathbb{R}\}\).

Step-by-step solution

Determine the general form of cross product for two given vectors

First, recall the formula for the cross product of two vectors \(\mathbf{a}=\langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b}=\langle b_1, b_2, b_3 \rangle\), which is given by: $$ \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle $$

Express the cross product of the given vectors

Now, let's denote the vector that we need to find as \(\mathbf{u}=\langle u_1, u_2, u_3 \rangle\). We have: $$ \langle 1,1,1\rangle \times \langle u_1, u_2, u_3\rangle = \langle 1u_3 - 1u_2, 1u_1 - 1u_3, 1u_2 - 1u_1\rangle = \langle u_3 - u_2, u_1 - u_3, u_2 - u_1\rangle $$

Set up the cross product equal to the given vector and solve for the components of \(\mathbf{u}\)

The cross product of the given vectors is equal to \(\langle -1, -1, 2\rangle\), so we can set up the following system of equations: $$ \begin{cases} u_3 - u_2 = -1 \\ u_1 - u_3 = -1 \\ u_2 - u_1 = 2 \end{cases} $$ To solve this system of equations, we can use substitution or elimination. From the first equation, we can find \(u_3\) in terms of \(u_2\): $$ u_3 = u_2 -1 $$ Now, substitute this into the second equation to find \(u_1\): $$ u_1 - (u_2 - 1) = -1 \\ u_1 = u_2 - 2 $$ Finally, substitute the expression for \(u_1\) into the third equation: $$ u_2 - (u_2 - 2) = 2 \\ 2 = 2 $$ As the third equation is always true, this indicates that there are infinitely many solutions for \(\mathbf{u}\).

Express \(\mathbf{u}\) as a function of a single variable

We can express \(\mathbf{u}\) in terms of just one variable, say, \(u_2\). Then, according to our previous calculations, we have: $$ \mathbf{u} = \langle u_1, u_2, u_3\rangle = \langle u_2 - 2, u_2, u_2 - 1 \rangle $$ Since \(u_2\) can be any real number, the set of all vectors \(\mathbf{u}\) that satisfy the given equation is: $$ \mathbf{u} = \{\langle u_2 - 2, u_2, u_2 - 1 \rangle \ | \ u_2 \in \mathbb{R}\} $$