find the angle between vectors <1,3> and <2,-5>
Answers
The angle
θ
between two vectors
→
A
and
→
B
is related to the modulus (or magnitude) and scaler (or dot) product of
→
A
and
→
B
by the relationship:
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→
A
⋅
→
B
=
|
A
|
|
B
|
cos
θ
By convention when we refer to the angle between vectors we choose the acute angle.
So for this problem, let the angle betwen
→
u
and
→
v
be
θ
then:
→
u
=
⟨
3
,
2
⟩
and
→
v
=
⟨
4
,
0
⟩
The angle is θ for the vector <1,3> and <2,-5> that is arc cos(-13/√290) by vector of and angle formula.
Given that,
We have 2 vectors <1,3> and <2,-5>
We have to find the angle between the vectors.
We know that,
Take a=<1,3> b=<2,5>
a.b = |a||b|cosθ (formula)
cosθ = (a.b)/|a||b|
|a| = √1²+3² = √1+9 = √10 (by modulus)
|b| = √2²+(-5)² = √5+25 = √29
a.b = 1×2+3×-5 = -13
cosθ = -13/√10×√29 (Substituting)
cosθ = -13/√290
θ = arc cos(-13/√290)
Therefore, The angle is θ for the vector <1,3> and <2,-5> that is arc cos(-13/√290) by vector of and angle formula.
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