Dot Product Calculating Angle

What is to be learned? How to use dot product to calculate the angle between vectors

From Before a.b = |a| |b| cosθ cosθ = a.b |a| |b| also a.b = x 1 x 2 + y 1 y 2 + z 1 z 2 x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b|

If vectors are perpendicular cosθ = cos90 0 = cosθ = x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b| 0 x 1 x 2 + y 1 y 2 + z 1 z 2 = 0 Numerator Denominator

cosθ = x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b| Find angle between a and b ( ) Numerator:2(7) + 1(2) + 4(6) = 40 Denom:|a| = √( ) = √21 |b| = √( ) = √89 cosθ = 40 √21√89 = 0.925θ =

Prove a = 3i – 2j is perpendicular to b = 4i + 6j - 5k cosθ = x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b| If perpendicular cosθ = cos90 0 = 0 i.e. num = 0 num = 3(4) + (-2)6 + 0(-5) = 0as required

Calculating Angles with Dot Product Rearranging formula cosθ = a.b |a| |b| x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b| = If perpendicular cosθ = cos90 0 = 0 i.e. num = 0

a = 3i + 4k,b = 4i + 6j + 2k cosθ = x 1 x 2 + y 1 y 2 + z 1 z 2 |a| |b| Calculate angle between vectors Numerator:3(4) + 0(6) + 4(2) = 20 Denom:|a| = √( ) = √25 = 5 |b| = √( ) = √56 cosθ = 20 5√56 = θ =

( ) 34k34k If a and b are perpendicular, find value of k a =b = If perpendicular cosθ = cos90 0 = 0 i.e. num = 0 num = 2(3) + (-1)4 + (-3)(k) = 2 – 3k 2 – 3k = 0 k = 2 / 3