Find the angle between two vectors?
p=3i+4j and q= i-2j
p=3i+4j and q= i-2j
Ian H
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
Como
-
p • q = ( 3i + 4j ) ( i - 2j )
p • q = 3 - 8
p • q = - 5
- 5 = 5 √5 cos ∅
cos ∅ = - 1/√5
∅ = 116•6 °
Engr. Ronald
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//
TomV
<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°