91影视

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 \(\mathbf{b}\).

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: $$ \langle 1u_3 - 1u_2, 1u_1 - 1u_3, 1u_2 - 1u_1 \rangle = \langle 0, 0, 1\rangle $$

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 \(\langle 1,1,1\rangle \times \mathbf{u} = \langle 0, 0, 1\rangle\).