ধরি, তিনটি ভেক্টর হলো: $$ \begin{aligned} \vec{A}&=A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \\ \vec{B}&=B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \\ \vec{C}&=C_x \hat{i} + C_y \hat{j} + C_z \hat{k} \end{aligned} $$ এখন, $$ \begin{aligned} \vec{B}+\vec{C}&=(B_x+C_x)\hat{i} + (B_y+C_y)\hat{j} + (B_z+C_z)\hat{k} \end{aligned} $$ বামপক্ষ, $$ \begin{aligned} \vec{A} \cdot (\vec{B}+\vec{C}) &= (A_x \hat{i} + A_y \hat{j} + A_z \hat{k}) \cdot [(B_x+C_x)\hat{i} + (B_y+C_y)\hat{j} + (B_z+C_z)\hat{k}] \\ &= A_x(B_x+C_x) + A_y(B_y+C_y) + A_z(B_z+C_z) \end{aligned} $$ ডানপক্ষ, $$ \begin{aligned} \vec{A} \cdot \vec{B} &= A_x B_x + A_y B_y + A_z B_z \\ \vec{A} \cdot \vec{C} &= A_x C_x + A_y C_y + A_z C_z \\ \therefore \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} &= (A_x B_x + A_y B_y + A_z B_z) + (A_x C_x + A_y C_y + A_z C_z) \\ &= A_x(B_x+C_x) + A_y(B_y+C_y) + A_z(B_z+C_z) \end{aligned} $$ সুতরাং, বামপক্ষ = ডানপক্ষ। অর্থাৎ, $$ \vec{A} \cdot (\vec{B}+\vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} $$ প্রমাণিত।