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Let \(\theta\) be the angle between a and b, where a and b are the vectors defined in \(a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}\). Find \(\cos^2 \theta\).

 Feb 28, 2019

Best Answer 

 #1
avatar+21842 
+3

Let \(\theta\) be the angle between a and b, where a and b are the vectors defined in

 

\(a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}\).
 

Find \(\cos^2 \theta\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\cos(\theta)} &\mathbf{=}& \mathbf{\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|} } \quad | \quad |\vec{a}| = \sqrt{2^2+(-6)^2+5^2}=\sqrt{65},\ |\vec{b}| = \sqrt{(-1)^2+(-2)^2+0^2}=\sqrt{5} \\\\ \cos(\theta) &=& \dfrac{\begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} } {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{2\cdot (-1)+(-6)\cdot(-2)+5\cdot 0} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{10} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos^2(\theta) &=& \dfrac{100} {65\cdot 5 } \\\\ \cos^2(\theta) &=& \dfrac{100} {325 } \\\\ \mathbf{\cos^2(\theta)} &\mathbf{=}& \mathbf{\dfrac{4}{13}} \\ \hline \end{array}\)

 

laugh

 Feb 28, 2019
 #1
avatar+21842 
+3
Best Answer

Let \(\theta\) be the angle between a and b, where a and b are the vectors defined in

 

\(a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}\).
 

Find \(\cos^2 \theta\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\cos(\theta)} &\mathbf{=}& \mathbf{\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|} } \quad | \quad |\vec{a}| = \sqrt{2^2+(-6)^2+5^2}=\sqrt{65},\ |\vec{b}| = \sqrt{(-1)^2+(-2)^2+0^2}=\sqrt{5} \\\\ \cos(\theta) &=& \dfrac{\begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} } {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{2\cdot (-1)+(-6)\cdot(-2)+5\cdot 0} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{10} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos^2(\theta) &=& \dfrac{100} {65\cdot 5 } \\\\ \cos^2(\theta) &=& \dfrac{100} {325 } \\\\ \mathbf{\cos^2(\theta)} &\mathbf{=}& \mathbf{\dfrac{4}{13}} \\ \hline \end{array}\)

 

laugh

heureka Feb 28, 2019

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