[ $0$ / $1$ Points] $\newline$ DETAILS $\newline$ PREVIOUS ANSWERS $\newline$ SPRECALC $\newline$ Find the angle between $\mathbf{u}$ and $\mathbf{v}$ , rounded to the nearest tenth degree. $\newline$ $\mathbf{u}=\langle 4,-4,-7\rangle, \mathbf{v}=\langle 7,4,4\rangle$

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1

Points]

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DETAILS

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PREVIOUS ANSWERS

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SPRECALC

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Find the angle between

\mathbf{u}

and

\mathbf{v}

, rounded to the nearest tenth degree.

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\mathbf{u}=\langle 4,-4,-7\rangle, \mathbf{v}=\langle 7,4,4\rangle

Dot Product Calculation: To find the angle between two vectors $u$ and $v$ , we can use the dot product formula, which is $u \cdot v = |u||v|\cos(\theta)$ , where $\theta$ is the angle between the vectors. We need to calculate the dot product of $u$ and $v$ , the magnitude of $u$ , and the magnitude of $v$ .
Magnitude of Vector u: First, let's calculate the dot product of vectors $u$ and $v$ . The dot product is given by $u \cdot v = u_1v_1 + u_2v_2 + u_3v_3$ , where $u_1, u_2, u_3$ are the components of vector $u$ and $v_1, v_2, v_3$ are the components of vector $v$ . So, for $u = (4, -4, -7)$ and $v = (7, 4, 4)$ , the dot product is $u \cdot v = (4)(7) + (-4)(4) + (-7)(4)$ .
Magnitude of Vector v: Calculating the dot product, we get $u \cdot v = 28 - 16 - 28 = -16$ .
Calculation of Cosine: Next, we need to find the magnitude of vector $u$ , which is $|u| = \sqrt{u_1^2 + u_2^2 + u_3^2}$ . For $u = (4, -4, -7)$ , this is $|u| = \sqrt{4^2 + (-4)^2 + (-7)^2}$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ .Calculating the magnitude of $v$ , we get $|v| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ .Calculating the magnitude of $v$ , we get $|v| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ .Now we have all the necessary components to find the angle $\theta$ . We can rearrange the dot product formula to solve for $\cos(\theta)$ : $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $0$ . Substituting the values we found, we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $1$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ .Calculating the magnitude of $v$ , we get $|v| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ .Now we have all the necessary components to find the angle $\theta$ . We can rearrange the dot product formula to solve for $\cos(\theta)$ : $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $0$ . Substituting the values we found, we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $1$ .Calculating $\cos(\theta)$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $3$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ . Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ . Calculating the magnitude of $v$ , we get $|v| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ . Now we have all the necessary components to find the angle $\theta$ . We can rearrange the dot product formula to solve for $\cos(\theta)$ : $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $0$ . Substituting the values we found, we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $1$ . Calculating $\cos(\theta)$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $3$ . To find the angle $\theta$ , we take the arccosine (inverse cosine) of $\cos(\theta)$ . So, $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $6$ .
Calculation of Angle: Calculating the magnitude of $u$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .Now, we need to find the magnitude of vector $v$ , which is $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For $v = (7, 4, 4)$ , this is $|v| = \sqrt{7^2 + 4^2 + 4^2}$ .Calculating the magnitude of $v$ , we get $|v| = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ .Now we have all the necessary components to find the angle $\theta$ . We can rearrange the dot product formula to solve for $\cos(\theta)$ : $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $0$ . Substituting the values we found, we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $1$ .Calculating $\cos(\theta)$ , we get $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $3$ .To find the angle $\theta$ , we take the arccosine (inverse cosine) of $\cos(\theta)$ . So, $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $6$ .Using a calculator, we find $|u| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$ $7$ degrees when rounded to the nearest tenth degree.