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# Let $\theta$ be the angle between $\bold{a}$ and $\bold{b}$, where $\bold{a}$ and $\bold{b}$ are the vectors defined in part (a). Find \$\cos

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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

#1
+23071
+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 $$\cos^2 \theta$$

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

Feb 28, 2019

#1
+23071
+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 $$\cos^2 \theta$$

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

heureka Feb 28, 2019