Find the angle between the given vectors to the nearest tenth of a degree. u = <- 5, 8>, v = <- 4, 8>

Solution:

Given: Vectors are u = -5 i + 8 j and v = -4i + 8 j

If θ is the angle between two vectors u and v, then

Cosθ = (u·v) / [|u| |v|]

Dot product of two vectors is u·v = a\(_1\) a\(_2\) + b\(_1\) b\(_2\)

u·v = (-5) (-4) + (8) (8) =20 + 64

∴ u.v = 84

|u| = √[(a\(_1\))² + (b\(_1\))²]

= √[(5)² + (8)²]

=√(25 + 64)

⇒ |u| = √(25 + 64) =√89

|v|= √[(a\(_2\) )² + (b\(_2\) )² ]

⇒ |v| =√[(-4)² + ( 8)²]

= √(16 + 64) = √80

|v|= √80

cosθ = 84/[√89 √80]

⇒ cosθ = 84/[9.443× 8.944]

⇒ cosθ = 84/84.38

⇒ cosθ = 0.9955

Therefore, θ = Cos^-1(0.995) = 5.4°

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Find the angle between the given vectors to the nearest tenth of a degree. u = <- 5, 8>, v = <- 4, 8>

Summary:

The angle between the given vectors to the nearest tenth of a degree. u = <- 5, 8>, v = <- 4, 8> is 5.4°.