Find the angle between two vectors?

Using analogous properties of complex vectors

If z = 3 + 4i then arg z = arctan(4/3), and,

if w = 1 – 2i then arg w = arctan(-2)

Required angle in radians is given by

arctan(4/3) - arctan(-2) ~ 0.927295 – (- 1.10715) = 2.03444

In degrees that is about 116.565

-

p • q = ( 3i + 4j ) ( i - 2j )

p • q = 3 - 8

p • q = - 5

- 5 = 5 √5 cos ∅

cos ∅ = - 1/√5

∅ = 116•6 °

p=3i+4j and q= i-2j

Find the dot product of the vectors.

p = (3,4) and q = (1, - 2)

p * q = p₁ q₁ + p₂ q₂

p * q = (3 * 1) + (4 * - 2)

p * q = 3 - 8

p * q = - 5

Find the magnitudes of each vector

||p|| = √(p₁^2 + p₂^2)

||p|| = √(3^2 + 4^2)

||p|| = 5

||q|| = √(q₁^2 + q₂^2)

||q|| = √[1^2 + (-2)^2]

||q|| = √(5)

Substitute and solve for θ

.....................p * q

cos(θ) = ----------------

................||p|| * ||q||

....................- 5

cos(θ) = ---------------- = -0.4472

............... (5) * √(5)

θ = 116.56° answer//

<p> = 3i + 4j

<q> = i - 2j

<p>•<q> = |p||q|cosΘ

cosΘ = (<p>•<q>/(|p||q|)

= (3•1 - 4•2)/√[(3²+4²)(1+2²)]

= -5/√125

= -√5/5

Ans: Θ = arccos(-√5/5) ≈ 116.57°