How do you compute the dot product for #<5, 12>*<-3, 2>#?

1 Answer
Nov 9, 2016

Please see the explanation of how to compute the dot-product for any two vectors of the same dimension and the answer to the given vectors: #< 5, 12 > * < -3, 2 > = 9#

Explanation:

For two dimensional vectors #< a_x, a_y > and < b_x, b_y >#, the dot-product is:

#< a_x, a_y > * < b_x, b_y > = (a_x)(b_x) + (a_y)(b_y)#

The dot-product is extendable to 3 dimensions and beyond:

For 3 dimensional vectors #< a_x, a_y, a_z > and < b_x, b_y, b_z >#, the dot-product is:

#< a_x, a_y, a_z > * < b_x, b_y, b_z > = (a_x)(b_x) + (a_y)(b_y) + (a_z)(b_z)#

For n dimensional vectors #< a_1, a_2,..., a_n > and < b_1, b_2,..., b_n >#, the dot-product is:

#< a_1, a_1,..., a_n > * < b_1, b_2,..., b_n > = (a_1)(b_1) + (a_2)(b_2) +..., + (a_n)(b_n)#

The dot product for the given vectors is:

#< 5, 12 > * < -3, 2 > = (5)(-3) + (12)(2)#

#< 5, 12 > * < -3, 2 > = -15 + 24#

#< 5, 12 > * < -3, 2 > = 9#