Let a=(-2,-5,0) and b=(-4,3,-1). Find the angle between the vector (in radius)

Answer 1

a=<-2,-5,0>

|a}= sqr(2^2+5^2 = sqr29

b =<-4,3,-1>

|b|= sqr(4^2 + 3^2 +1^2) = sqr26

|a||b|= sqr(29x26) = sqr754

dot product = 8-15 = -7

Cosine of the Angle = dot product divideded by the product of the

magnitudes

CosT = -7/sqr754

T = Cos^-1(-7/sqr754)

the angle = the inverse cosine of -7/sqr754

at this point use a calculator to get an approximation

if you get degress, then convert it to radians

Answer 2

a·b=¦a¦ ¦b¦cos(θ)

a=(-2,-5,0), b=(-4,3,-1)

¦a¦=((-2)² + (-5)² + 0²)^0.5 or 5.385

¦b¦=((-4)² + 3² + (-1)²)^0.5 or ·5.099

a·b=-2×-4 + -5×3 + 0×-1 or -7

cos(θ)=-7/(5.385×5.099)

θ=1.83 radians