Chapter 9.5, Problem 34AYU

To determine

To prove: Distributive property $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$.

Answer to Problem 34AYU

Solution:

$u\cdot \left(v+w\right)=u\cdot v+u\cdot w$

Explanation of Solution

Let $u=\langle a_1,b_1\rangle ,v=\langle a_2,b_2\rangle$ and $w=\langle a_3,b_3\rangle$. Then,

$u\cdot \left(v+w\right)=\langle a_1,b_1\rangle \cdot \left(\langle a_2,b_2\rangle +\langle a_3,b_3\rangle \right)$

$=\langle a_1,b_1\rangle \cdot \langle a_2+a_3,b_2+b_3\rangle$

$=a_1\left(a_2+a_3\right)+b_1\left(b_2+b_3\right)$

$=a_1{a}_2+a_1{a}_3+b_1{b}_2+b_1{b}_3$

$=\left(a_1{a}_2+b_1{b}_2\right)+\left(a_1{a}_3+b_1{b}_3\right)$

$=u\cdot v+u\cdot w$

$u\cdot \left(v+w\right)=u\cdot v+u\cdot w$