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

Question

8961 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-02T14:33:52

#= vec 0#

this is the same as

#<1,0,0,>.<0,1,1> #

#= 1*0 + 0 * 1 + 0*1 #

#= vec 0#

you can also use the distributive law

#\vec {A} * (\vec {B} +\vec {C} )=\vec {A} * \vec {B} +\vec {A} * \vec {C} #

so

#i*(j+k)#
#= i*j + i* k#
# = 0 + 0 #