91Ó°ÊÓ

The cross product is distributive if the prohection of sum of vectors on the former vector is additive.

In the following diagram, the vectors ${\mathrm{B}}_{\mathrm{A}}$and $C_A$are the projection of vectors $B$and $C$ on the vector A, such that total projection on the vector A is ${\mathrm{B}}_{\mathrm{A}}+{\mathrm{C}}_{\mathrm{A}}$

Now obtain the projection of vector$B$and $C$ on the vector A.

$\left|\left|{\mathbf{B}}_A\right|\right|=\frac{\mathbf{B}â‹\dots \mathbf{A}}{\left|\left|\mathbf{A}\right|\right|}$ ……. (1)

Solve for the vector $\left|\left|{\mathbf{C}}_A\right|\right|$.

$\left|\left|{\mathbf{C}}_A\right|\right|=\frac{\mathbf{C}â‹\dots \mathbf{A}}{\left|\left|\mathbf{A}\right|\right|}$ …â¶Ä¦â¶Ä¦.(2)

Add equation (1) and (2)

$\left|\left|{\mathrm{B}}_{\mathrm{A}}\right|\right|+\left|\left|{\mathrm{C}}_{\mathrm{A}}\right|\right|\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=\frac{\mathrm{B}â‹\dots \mathrm{A}}{\left|\left|\mathrm{A}\right|\right|}\text{ â¶Ä‰â¶Ä‰}+\frac{\mathrm{C}â‹\dots \mathrm{A}}{\left|\left|\mathrm{A}\right|\right|}$ …â¶Ä¦â¶Ä¦â¶Ä¦.(3)

Now obtain the total projection on vector $A$.

$\begin{array}{c}\left||{\mathrm{B}}_{\mathrm{A}}+{\mathrm{C}}_{\mathrm{A}}|\right|=\frac{(\mathrm{B}+\mathrm{C})â‹\dots \mathrm{A}}{\left|\left|\mathrm{A}\right|\right|}\\ =\frac{\mathrm{B}â‹\dots \mathrm{A}}{\left|\left|\mathrm{A}\right|\right|}\text{ â¶Ä‰â¶Ä‰}+\frac{\mathrm{C}â‹\dots \mathrm{A}}{\left|\left|\mathrm{A}\right|\right|}\end{array}$ …â¶Ä¦..(4)

From the equation (3) and (4), $\left|\left|B_A\right|\right|+\left|\left|C_A\right|\right|\text{ }=\left||B_A+C_A|\right|$. Thus, the dot product $A×(B+C)$is distributive .similarly the dot product $\mathrm{A}â‹\dots ≤(\mathrm{B}+\mathrm{C})$is also distributive.

The dot product is distributive if the dot product of former vector with each vector in sum of vectors is added to give resultant dot product.

Let the vectors are defined as $\mathrm{A}={\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k}$, $B=B_{x}i+B_{y}j+B_{z}k$ and $C=C_{x}i+C_{y}j+C_{z}k$.

Evaluate the value of $Aâ‹\dots (B+C)$.

$\begin{array}{c}\mathrm{A}â‹\dots (\mathrm{B}+\mathrm{C})=({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k})(({\mathrm{B}}_{\mathrm{x}}+{\mathrm{C}}_{\mathrm{x}})\mathrm{i}+({\mathrm{B}}_{\mathrm{y}}+{\mathrm{C}}_{\mathrm{y}})\mathrm{j}+\left(({\mathrm{B}}_{\mathrm{z}}+{\mathrm{C}}_{\mathrm{z}})\right)\mathrm{k})\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}=\text{ }({\mathrm{B}}_{\mathrm{x}}+{\mathrm{C}}_{\mathrm{x}}){\mathrm{A}}_{\mathrm{x}}+({\mathrm{B}}_{\mathrm{y}}+{\mathrm{C}}_{\mathrm{y}}){\mathrm{A}}_{\mathrm{y}}+\left(({\mathrm{B}}_{\mathrm{z}}+{\mathrm{C}}_{\mathrm{z}})\right){\mathrm{A}}_{\mathrm{z}}\end{array}$ …(5)

Now, evaluate $(Aâ‹\dots B)+(Aâ‹\dots C)$.

$\begin{array}{c}(\mathrm{A}â‹\dots \mathrm{B})+(\mathrm{A}â‹\dots \mathrm{C})=({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k})â‹\dots ({\mathrm{B}}_{\mathrm{x}}\mathrm{i}+{\mathrm{B}}_{\mathrm{y}}\mathrm{j}+{\mathrm{B}}_{\mathrm{z}}\mathrm{k})+({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k})â‹\dots ({\mathrm{C}}_{\mathrm{x}}\mathrm{i}+{\mathrm{C}}_{\mathrm{y}}\mathrm{j}+{\mathrm{C}}_{\mathrm{z}}\mathrm{k})\\ \text{ }=\text{ }({\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{x}}+{\mathrm{A}}_{\mathrm{x}}{\mathrm{C}}_{\mathrm{x}})+({\mathrm{A}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{A}}_{\mathrm{y}}{\mathrm{C}}_{\mathrm{y}})+({\mathrm{A}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{z}}+{\mathrm{A}}_{\mathrm{z}}{\mathrm{C}}_{\mathrm{z}})\\ =({\mathrm{B}}_{\mathrm{x}}+{\mathrm{C}}_{\mathrm{x}}){\mathrm{A}}_{\mathrm{x}}+({\mathrm{B}}_{\mathrm{y}}+{\mathrm{C}}_{\mathrm{y}}){\mathrm{A}}_{\mathrm{y}}+({\mathrm{B}}_{\mathrm{z}}+{\mathrm{C}}_{\mathrm{z}}){\mathrm{A}}_{\mathrm{z}}\end{array}$…â¶Ä¦(6)

From equation (5) and (6), it can be concluded that the result of $(Aâ‹\dots B)+(Aâ‹\dots C)$ and $Aâ‹\dots (B+C)$ is same.

Thus, $Aâ‹\dots (B+C)=(Aâ‹\dots B)+(Aâ‹\dots C)$, which is distributive.