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To determine the angle [tex]\(\theta\)[/tex] between the vectors [tex]\(\vec{C} = 2 \hat{i} + 7 \hat{j}\)[/tex] and [tex]\(\vec{D} = 14 \hat{i} + 38 \hat{j}\)[/tex], follow these steps:
1. Find the dot product [tex]\(\vec{C} \cdot \vec{D}\)[/tex]:
The dot product of two vectors [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex] is given by:
[tex]\[ \vec{C} \cdot \vec{D} = C_x \cdot D_x + C_y \cdot D_y \][/tex]
For [tex]\(\vec{C} = \begin{pmatrix}2 \\ 7\end{pmatrix}\)[/tex] and [tex]\(\vec{D} = \begin{pmatrix}14 \\ 38\end{pmatrix}\)[/tex], the dot product is:
[tex]\[ \vec{C} \cdot \vec{D} = (2 \cdot 14) + (7 \cdot 38) = 28 + 266 = 294 \][/tex]
2. Calculate the magnitudes of [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex]:
The magnitude of a vector [tex]\(\vec{V} = \begin{pmatrix}V_x \\ V_y\end{pmatrix}\)[/tex] is given by:
[tex]\[ \|\vec{V}\| = \sqrt{V_x^2 + V_y^2} \][/tex]
For [tex]\(\vec{C} = \begin{pmatrix}2 \\ 7\end{pmatrix}\)[/tex]:
[tex]\[ \|\vec{C}\| = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.280 \][/tex]
For [tex]\(\vec{D} = \begin{pmatrix}14 \\ 38\end{pmatrix}\)[/tex]:
[tex]\[ \|\vec{D}\| = \sqrt{14^2 + 38^2} = \sqrt{196 + 1444} = \sqrt{1640} \approx 40.497 \][/tex]
3. Determine the cosine of the angle [tex]\(\theta\)[/tex]:
The cosine of the angle [tex]\(\theta\)[/tex] between two vectors [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex] is given by:
[tex]\[ \cos \theta = \frac{\vec{C} \cdot \vec{D}}{\|\vec{C}\| \|\vec{D}\|} \][/tex]
Substituting the values found:
[tex]\[ \cos \theta = \frac{294}{7.280 \times 40.497} \approx \frac{294}{294.726} \approx 0.997 \][/tex]
4. Find the angle [tex]\(\theta\)[/tex]:
The angle [tex]\(\theta\)[/tex] can be found by taking the inverse cosine (arccos) of the cosine value:
[tex]\[ \theta = \arccos(0.997) \approx 0.075 \text{ radians} \][/tex]
5. Convert the angle from radians to degrees:
Since [tex]\(1 \text{ radian} \approx 57.2958 \text{ degrees}\)[/tex]:
[tex]\[ \theta \approx 0.075 \text{ radians} \times 57.2958 \text{ degrees/radian} \approx 4.28 \text{ degrees} \][/tex]
Given this result, the closest match from the provided options is:
b. [tex]\(\theta = 4^\circ\)[/tex]
1. Find the dot product [tex]\(\vec{C} \cdot \vec{D}\)[/tex]:
The dot product of two vectors [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex] is given by:
[tex]\[ \vec{C} \cdot \vec{D} = C_x \cdot D_x + C_y \cdot D_y \][/tex]
For [tex]\(\vec{C} = \begin{pmatrix}2 \\ 7\end{pmatrix}\)[/tex] and [tex]\(\vec{D} = \begin{pmatrix}14 \\ 38\end{pmatrix}\)[/tex], the dot product is:
[tex]\[ \vec{C} \cdot \vec{D} = (2 \cdot 14) + (7 \cdot 38) = 28 + 266 = 294 \][/tex]
2. Calculate the magnitudes of [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex]:
The magnitude of a vector [tex]\(\vec{V} = \begin{pmatrix}V_x \\ V_y\end{pmatrix}\)[/tex] is given by:
[tex]\[ \|\vec{V}\| = \sqrt{V_x^2 + V_y^2} \][/tex]
For [tex]\(\vec{C} = \begin{pmatrix}2 \\ 7\end{pmatrix}\)[/tex]:
[tex]\[ \|\vec{C}\| = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.280 \][/tex]
For [tex]\(\vec{D} = \begin{pmatrix}14 \\ 38\end{pmatrix}\)[/tex]:
[tex]\[ \|\vec{D}\| = \sqrt{14^2 + 38^2} = \sqrt{196 + 1444} = \sqrt{1640} \approx 40.497 \][/tex]
3. Determine the cosine of the angle [tex]\(\theta\)[/tex]:
The cosine of the angle [tex]\(\theta\)[/tex] between two vectors [tex]\(\vec{C}\)[/tex] and [tex]\(\vec{D}\)[/tex] is given by:
[tex]\[ \cos \theta = \frac{\vec{C} \cdot \vec{D}}{\|\vec{C}\| \|\vec{D}\|} \][/tex]
Substituting the values found:
[tex]\[ \cos \theta = \frac{294}{7.280 \times 40.497} \approx \frac{294}{294.726} \approx 0.997 \][/tex]
4. Find the angle [tex]\(\theta\)[/tex]:
The angle [tex]\(\theta\)[/tex] can be found by taking the inverse cosine (arccos) of the cosine value:
[tex]\[ \theta = \arccos(0.997) \approx 0.075 \text{ radians} \][/tex]
5. Convert the angle from radians to degrees:
Since [tex]\(1 \text{ radian} \approx 57.2958 \text{ degrees}\)[/tex]:
[tex]\[ \theta \approx 0.075 \text{ radians} \times 57.2958 \text{ degrees/radian} \approx 4.28 \text{ degrees} \][/tex]
Given this result, the closest match from the provided options is:
b. [tex]\(\theta = 4^\circ\)[/tex]
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