Problem on proving that dot products are distributive

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Prove for 2d vectors: $\bfa \cdot (\bfb + \bfc) = \bfa\cdot \bfb + \bfa \cdot \bfc.$

  • Solution

    To prove this identity, we appeal to the componentwise definitions of dot product and addition. We evaluate the left hand side and the right hand side in terms of their components. If the results are equal, the identity is true.
    Recall that
    The vector sum of $\langle x_1, x_2 \rangle + \langle y_1, y_2\rangle = \langle x_1 + y_1, x_2+ y_2 \rangle.$
    The dot product is defined as $\langle x_1, x_2 \rangle \cdot \langle y_1, y_2\rangle = x_1 y_1 + x_2 y_2.$
    Let's introduce components of $\bfa, \bfb, \bfc$:\begin{align}
    \bfa &= \langle a_1, a_2 \rangle\\
    \bfb &= \langle b_1, b_2 \rangle\\
    \bfc &= \langle c_1, c_2 \rangle
    \end{align}

    Left hand side

    We now evaluate $\bfa \cdot (\bfb + \bfc)$ in terms of components.
    The sum $\bfb + \bfc$ is given by $$ \bfb + \bfc = \langle b_1+c_1, b_2+c_2 \rangle.$$
    The dot product $\bfa \cdot (\bfb + \bfc)$ is then given by \begin{align}\langle a_1, a_2 \rangle \cdot \langle b_1+c_1, b_2+c_2\rangle &= a_1(b_1+c_1) + a_2 (b_2 + c_2)\\
    &= a_1 b_1 + a_1 c_1 + a_2 b_2 + a_2 c_2.\end{align}

    Right hand side

    We now evaluate $\bfa\cdot \bfb + \bfb \cdot \bfc$.
    We compute that $\bfa\cdot\bfb$ is $$\langle a_1, a_2 \rangle \cdot \langle b_1, b_2 \rangle =a_1 b_1 + a_2 b_2.$$
    We compute that $\bfa\cdot\bfc$ is $$\langle a_1, a_2 \rangle \cdot \langle c_1, c_2 \rangle =a_1 c_1 + a_2 c_2.$$
    The sum is thus $$\bfa \cdot \bfb + \bfb \cdot \bfc = a_1 b_1 + a_1 c_1 + a_2 b_2 + a_2 c_2.$$

    Conclusion

    We observe that the left hand and right hand sides of the expression are equal for all 2d vectors $\bfa, \bfb, \bfc$. We conclude $$\bfa \cdot (\bfb + \bfc) = \bfa\cdot \bfb + \bfa \cdot \bfc.$$

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