# How do you compute the dot product for i*(j+k)?

##### 1 Answer
Jul 2, 2016

$= \vec{0}$

#### Explanation:

this is the same as

$< 1 , 0 , 0 , > . < 0 , 1 , 1 >$

$= 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 1$

$= \vec{0}$

you can also use the distributive law

$\setminus \vec{A} \cdot \left(\setminus \vec{B} + \setminus \vec{C}\right) = \setminus \vec{A} \cdot \setminus \vec{B} + \setminus \vec{A} \cdot \setminus \vec{C}$

so

$i \cdot \left(j + k\right)$
$= i \cdot j + i \cdot k$
$= 0 + 0$