# What is scalar product of a=2i-3j, b=6i+4j?

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Remember the formula of dot scalar product:

`bar a` * `barb` = |bara|*|barb|*cos theta

theta is the angle between bara and barb

|a| is the magnitude of the vector a: bara = sqrt(2^2 + (-3)^2)

bara = sqrt 13

|b| is the magnitude of the vector b:

`barb = sqrt(6^2 + 4^2)` `bara = sqrt 52 = 2sqrt13`

`` `bara*barb = (sqrt13)*(2sqrt13) cos theta`

**The scalar product is `bara*barb` =26 cos `theta` .**