With the cross product of two vectors defined by (4.14), show that finding the cross product is a linear operation, that is, show that (4.18) is valid. Warning hint: Don't try to prove it by writing out components: Writing, for example, ${\mathbf{iA}}_{\mathbf{x}}\mathbf{}\mathbf{x}\left({\mathrm{jB}}_x+{\mathrm{kB}}_z\right)\mathbf{=}{\mathbf{iA}}_{\mathbf{x}}\mathbf{}\mathbf{}{\mathbf{xjB}}_{\mathbf{y}}\mathbf{}\mathbf{+}{\mathbf{iA}}_{\mathbf{x}}\mathbf{}{\mathbf{xkB}}_{\mathbf{z}}$would be assuming what you're trying to prove. Further hints: First show that (4.18) is valid if $\mathbf{B}$ and $\mathbf{C}$ are both perpendicular to Aby sketching (in the plane perpendicular to A ) the vectors B,C,B+C, and their vector products with A. Then do the general case by first showing that $\mathbf{A}\mathbf{\times }\mathbf{B}$ and $\mathbf{A}\mathbf{\times }{\mathbf{B}}_{\mathbf{\perp }}$(where ${\mathbf{B}}_{\mathbf{\perp }}$is the vector component ofBperpendicular to A ) have the same magnitude and the same direction.

In three-dimensional space, the cross product is a binary operation between two vectors. It produces a perpendicular vector to both vectors. a b represents the vector product of two vectors a and b.

The cross-product of two vectors needs to be a linear operation.

Consider the general vectors.

$$\begin{gathered} \mathrm{a}=\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{x}}\\ {\mathrm{a}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}\end{array}\right] \\ \mathrm{b}=\left[\begin{array}{c}{\mathrm{b}}_{\mathrm{x}}\\ {\mathrm{b}}_{\mathrm{y}}\\ {\mathrm{b}}_{\mathrm{z}}\end{array}\right] \\ \mathrm{c}=\left[\begin{array}{c}{\mathrm{c}}_{\mathrm{x}}\\ {\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{c}}_{\mathrm{z}}\end{array}\right] \end{gathered}$$

Find the matrix b + c .

$$\begin{gathered} \mathrm{b}+\mathrm{c}=\left[\begin{array}{c}{\mathrm{b}}_{\mathrm{x}}\\ {\mathrm{b}}_{\mathrm{y}}\\ {\mathrm{b}}_{\mathrm{z}}\end{array}\right]+\left[\begin{array}{c}{\mathrm{c}}_{\mathrm{x}}\\ {\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{c}}_{\mathrm{z}}\end{array}\right] \\ =\left[\begin{array}{c}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{c}}_{\mathrm{x}}\\ {\mathrm{b}}_{\mathrm{y}}+{\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{b}}_{\mathrm{z}}+{\mathrm{c}}_{\mathrm{z}}\end{array}\right] \end{gathered}$$

Compute the cross-product of and, .

role="math" localid="1658996008489" $$\begin{gathered} \mathrm{a}\times \left(\mathrm{b}+\mathrm{c}\right)=\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{y}}\left({\mathrm{b}}_{\mathrm{z}}+{\mathrm{c}}_{\mathrm{z}}\right)-{\mathrm{a}}_{\mathrm{z}}\left({\mathrm{b}}_{\mathrm{y}}+{\mathrm{c}}_{\mathrm{y}}\right)\\ {\mathrm{a}}_{\mathrm{z}}\left({\mathrm{b}}_{\mathrm{x}}+{\mathrm{c}}_{\mathrm{x}}\right)-{\mathrm{a}}_{\mathrm{x}}\left({\mathrm{b}}_{\mathrm{z}}+{\mathrm{c}}_{\mathrm{z}}\right)\\ {\mathrm{a}}_{\mathrm{x}}\left({\mathrm{b}}_{\mathrm{y}}+{\mathrm{c}}_{\mathrm{y}}\right)-{\mathrm{a}}_{\mathrm{y}}\left({\mathrm{b}}_{\mathrm{x}}+{\mathrm{c}}_{\mathrm{x}}\right)\end{array}\right] \\ =\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{z}}\\ {\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{y}}+{\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{x}}\end{array}\right] \\ =\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{y}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{z}}+{\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{z}}\\ {\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{x}}\end{array}\right] \\ =\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{z}}\\ {\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{x}}\end{array}\right]+\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{z}}-{\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{x}}-{\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{z}}\\ {\mathrm{a}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{y}}-{\mathrm{a}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{x}}\end{array}\right] \end{gathered}$$

Further, solve.

$$\begin{gathered} \mathrm{a}\times \left(\mathrm{b}+\mathrm{c}\right)=\left(\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{x}}\\ {\mathrm{a}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}\end{array}\right]\times \left[\begin{array}{c}{\mathrm{b}}_{\mathrm{x}}\\ {\mathrm{b}}_{\mathrm{y}}\\ {\mathrm{b}}_{\mathrm{z}}\end{array}\right]\right)+\left(\left[\begin{array}{c}{\mathrm{a}}_{\mathrm{x}}\\ {\mathrm{a}}_{\mathrm{y}}\\ {\mathrm{a}}_{\mathrm{z}}\end{array}\right]\times \left[\begin{array}{c}{\mathrm{c}}_{\mathrm{x}}\\ {\mathrm{c}}_{\mathrm{y}}\\ {\mathrm{c}}_{\mathrm{z}}\end{array}\right]\right) \\ =\left(\mathrm{a}\times \mathrm{b}\right)+\left(\mathrm{a}\times \mathrm{c}\right) \end{gathered}$$

Therefore, the desired relationship is proved.