91Ó°ÊÓ

The cross product of a vector with itself is always the zero vector. This is because the cross product is calculated as:\[\mathbf{u} \times \mathbf{u} = (u_2 u_3 - u_3 u_2) \hat{\mathbf{i}} + (u_3 u_1 - u_1 u_3) \hat{\mathbf{j}} + (u_1 u_2 - u_2 u_1) \hat{\mathbf{k}}.\]Each component in the cross product involves terms like \((u_i u_i) - (u_i u_i)\), which always equal zero. Therefore, \(\mathbf{u} \times \mathbf{u} = \mathbf{0}\).

This property shows the distributive nature of the cross product over vector addition. The cross product follows the following rule:\[ \mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ u_1 & u_2 & u_3 \ v_1 + w_1 & v_2 + w_2 & v_3 + w_3 \end{vmatrix}.\]Expanding the determinant, you'll find the expression results in:\[ (u_2 (v_3 + w_3) - u_3 (v_2 + w_2)) \hat{\mathbf{i}} + (u_3 (v_1 + w_1) - u_1 (v_3 + w_3)) \hat{\mathbf{j}} + (u_1 (v_2 + w_2) - u_2 (v_1 + w_1)) \hat{\mathbf{k}}.\]This expands to:\[ (u_2 v_3 - u_3 v_2 + u_2 w_3 - u_3 w_2) \hat{\mathbf{i}} + (u_3 v_1 - u_1 v_3 + u_3 w_1 - u_1 w_3) \hat{\mathbf{j}} + (u_1 v_2 - u_2 v_1 + u_1 w_2 - u_2 w_1) \hat{\mathbf{k}}.\]Which confirms:\[ (\mathbf{u} \times \mathbf{v}) + (\mathbf{u} \times \mathbf{w}). \]

This property expresses that scalar multiplication can be factored out of the cross product. Let's prove the first equality: \[ c(\mathbf{u} \times \mathbf{v}) = c \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}, \]which is equivalent to:\[ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ cu_1 & cu_2 & cu_3 \ v_1 & v_2 & v_3 \end{vmatrix} = (c \mathbf{u}) \times \mathbf{v}. \]The calculation follows similarly for \( \mathbf{u} \times (c \mathbf{v}) \):\[ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ u_1 & u_2 & u_3 \ cv_1 & cv_2 & cv_3 \end{vmatrix}, \]proving that scalar values can distribute in front of either vector.

This property illustrates that the vector \( \mathbf{u} \) is orthogonal to the cross product \( \mathbf{u} \times \mathbf{v} \). By the definition of the dot product:\[ \mathbf{u} \cdot(\mathbf{u} \times \mathbf{v}) = u_1 (u_2 v_3 - u_3 v_2) + u_2 (u_3 v_1 - u_1 v_3) + u_3 (u_1 v_2 - u_2 v_1). \]Simplifying this expression by expanding each component shows that they sum to zero:\[ (u_1 u_2 v_3 - u_1 u_3 v_2) + (u_2 u_3 v_1 - u_2 u_1 v_3) + (u_3 u_1 v_2 - u_3 u_2 v_1) = 0. \]Thus, we confirm that \( \mathbf{u} \cdot(\mathbf{u} \times \mathbf{v}) = \mathbf{0} \).