Given vectors u = ⟨2, –3⟩ and v = ⟨1, –1⟩, what is the measure of the angle between the vectors?

Answers

Answer 1

Answer:

The measure of the angle between the vectors = Ф = 11.30°

Step-by-step explanation:

Given

u = ⟨2, –3⟩v = ⟨1, –1⟩

[tex]\mathrm{Computing\:the\:angle\:between\:the\:vectors}:\quad \cos \left(\theta \right)\:=\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|\cdot \left|\vec{b\:}\right|}[/tex]

Next, find the lengths of the vectors:

[tex]\mathrm{Computing\:the\:Euclidean\:Length\:of\:a\:vector}:\quad \left|\left(x_1\:,\:\:\ldots \:,\:\:x_n\right)\right|=\sqrt{\sum _{i=1}^n\left|x_i\right|^2}[/tex]

u = ⟨2, –3⟩

[tex]\:\:\left|u\right|\:=\sqrt{2^2+\left(-3\right)^2}[/tex]

     [tex]=\sqrt{13}[/tex]

u = ⟨2, –3⟩

[tex]|v|=\sqrt{1^2+\left(-1\right)^2}[/tex]

    [tex]=\sqrt{2}[/tex]

Finally, the angle is given by:

[tex]\mathrm{Computing\:the\:angle\:between\:the\:vectors}:\quad \cos \left(\theta \right)\:=\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|\cdot \left|\vec{b\:}\right|}[/tex]

cos (Ф) = 5/√26

Ф = arc cos (cos (Ф)) = arc cos (5 √26) / (26)

Ф = 11.30°

Thus, the measure of the angle between the vectors = Ф = 11.30°

Answer 2

Answer:

The Answer is A. 11.3

Step-by-step explanation:

got it right. Also thats just the letter answer  :)


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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Hoped I helped Im Eve btw. Have an amazing day and consider marking this as brainliest if you do thank you in advanced! :3

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Answers

Answer:

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Step-by-step explanation:

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دائرة مركزها o فيها الوتران cd,ab اذا علمت ان النقطة n نقطة منتصف الوتر cd والنقطة m نقطة منتصف الوترab وكان om=on وطول cn=6cm
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Step-by-step explanation:

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Answers

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Step-by-step explanation:

Answer:

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Answers

Answer:

Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Answers

Answer:

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Step-by-step explanation:

To solve this problem, we can make the width as x.

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