Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, 4>, v = <7, 5>

Solution:

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

cosθ = (u·v) / [lul lvl]

Given u = 6 i + 4 j

v = 7i + 5 j

Dot product of two vectors is u·v = a₁a₂ + b₁b₂

u·v = (6)(7) + (4)(5) = 42 + 20

Therefore,

u.v = 62

llull = √[(a₁)² + (b₁)² ]

= √[(6)² + (4)²]

=√36 + 16

lul = √36 + 16 = √52

lvl = √[(a₂)² + (b₂)²]

= √[(7)² + (5)²] 

lvl = √49 + 25 = √74,

cos θ = 62 / [√52 √74]

⇒ cosθ = 62 / [√52 × √74]

⇒ cosθ = 62 / √3848

θ = cos^-1(0.999) = 0.9

Hence, nearest tenth will be 1 degree.

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Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, 4>, v = <7, 5>

Summary:

The angle between the given vectors to the nearest tenth of a degree.u = <6, 4>, v = <7, 5> to the nearest tenth of a degree is 1 degree.