Coplanarity Condition via Scalar Triple Product
Given three vectors \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}\), and \(\mathbf{c} = \begin{pmatrix} k \\ 2 \\ 1 \end{pmatrix}\), determine the value of \(k\) such that the vectors are coplanar.
Part a
Compute the scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\) where \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}\), and \(\mathbf{c} = \begin{pmatrix} k \\ 2 \\ 1 \end{pmatrix}\).
Exploring Cross Product Properties
Let \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix}\) and \(\mathbf{c} = \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}\). (a) Show that the cross product is distributive over vector addition, that is, demonstrate that \(\mathbf{a} \times (\mathbf{b}+\mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\). […]
Determining Cross Product Direction via Right-Hand Rule
Consider the vectors \(\mathbf{a} = (2, -3, 4)\) and \(\mathbf{b} = (-1, 5, 2)\). Using the right-hand rule, determine the direction of the vector product \(\mathbf{a} \times \mathbf{b}\). Provide a brief explanation of your reasoning.