Dot products?

Question

a = -4 i + 4 j and b = -6 i

I would like to find out how to work out a.b (a dot b)and find angle between the vectors in degrees to one decimal place

Thanks for any help

Answer 1

a.b = -4*(-6) + 4* 0 = -24

mod (a) = sqrt((-4)^2+4^2) = sqrt(32)
mod b = 6

cos (theta) = (a.b)/(mod (a).mod(b)) = -24/(6*4*(sqrt(2)))

= -6*sqrt(2)/12 = -1/sqrt(2) so theta = 3pi/4 ot 5pi/4

Answer 2

Think of b as being -6i + 0j
a dot b will be -4*-6 + 4*0 = 24
a dot b = |a| * |b| * cos (theta)

I already showed you what a dot b is.
Find the magnitude of a and b.
solve the equation for theta and you have your answer.

Answer 3

a=-4i+4j and b= -6i+0j

a.b = -4*(-6) + 4* 0 = -24

mod (a) = sqrt((-4)^2+4^2) = sqrt(32)
mod b = 6

cos (theta) = (a.b)/(mod (a).mod(b)) = -24/(6*(sqrt(32)))

= -24/(6*4*sqrt(2)) = -1/sqrt(2) = 0.7071

theta = cos inverse of -(0.7071) = 45 degrees.