Find the angle between the vectors &lt;1,3&gt;and&lt;−2,4&gt;

Remember that the angle between two vectors u and v is defined by cos⁡(θ)=u⋅v||u||||v|| So lets calculate u⋅v,||u||,and||v||foru=⟨1,3⟩andv=⟨−2,4⟩ u⋅v=⟨1,3⟩⋅⟨−2,4⟩=−2+12=10 ||u||=12+32=10 ||v||=(−2)2+42=20 Now plugging these into our equation, we find that cos⁡(θ)=101020=12 Looking at the unit circle, we see that cos⁡(θ)=12 at 45∘and315∘. Because the angle between two vectors must be between 0∘and180∘, that means that the angle between vectors u and v is 45∘

Expert answers to "Find the angle between the vectors <1,3>and<−2,4>"

College
Algebra
Linear algebra
Vectors and spaces

facas9

Answered question

2021-01-02

Find the angle between the vectors

< 1, 3 > and < - 2, 4 >

Answer & Explanation

Jozlyn

Skilled2021-01-03Added 85 answers

Remember that the angle between two vectors u and v is defined by

\cos (θ) = \frac{u \cdot v}{| | u | | | | v | |}

So lets calculate

u \cdot v, | | u | |, and | | v | | f or u = ⟨ 1, 3 ⟩ and v = ⟨ - 2, 4 ⟩

u \cdot v = ⟨ 1, 3 ⟩ \cdot ⟨ - 2, 4 ⟩ = - 2 + 12 = 10

| | u | | = \sqrt{1^{2} + 3^{2}} = \sqrt{10}

| | v | | = \sqrt{{(- 2)}^{2} + 4^{2}} = \sqrt{20}

Now plugging these into our equation, we find that

\cos (θ) = \frac{10}{\sqrt{10} \sqrt{20}} = \frac{1}{\sqrt{2}}

Looking at the unit circle, we see that

\cos (θ) = \frac{1}{\sqrt{2}} a t 45^{\circ} and 315^{\circ}

. Because the angle between two vectors must be between

0^{\circ} and 180^{\circ}

, that means that the angle between vectors u and v is

45^{\circ}

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