What is most nearly the degreees in acute angle 321 232 vector

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what is most nearly the degreees in acute angle 321 232 vector

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LectureNotes said what is most nearly the degrees in the acute angle between the vectors (3, 2, 1) and (2, 3, 2)?

Answer:
The acute angle between the two vectors (3, 2, 1) and (2, 3, 2) is approximately 24.7°.

To find this angle, we can use the dot product formula and vector magnitudes:

  1. Dot product:

    \mathbf{v_1} \cdot \mathbf{v_2} = 3 \times 2 + 2 \times 3 + 1 \times 2 = 6 + 6 + 2 = 14

  2. Magnitudes:

    \|\mathbf{v_1}\| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}, \quad \|\mathbf{v_2}\| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{17}

  3. Cosine of angle:

    \cos \theta = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{\|\mathbf{v_1}\|\|\mathbf{v_2}\|} = \frac{14}{\sqrt{14}\sqrt{17}} \approx 0.908

  4. Angle (in degrees):

    \theta = \cos^{-1}(0.908) \approx 24.7^\circ

Calculation Summary Table

Step Expression/Value Result
1. Dot Product (3)(2) + (2)(3) + (1)(2) 14
2. ‖v₁‖ \sqrt{3^2 + 2^2 + 1^2} \sqrt{14}
3. ‖v₂‖ \sqrt{2^2 + 3^2 + 2^2} \sqrt{17}
4. Cosine of Angle \frac{14}{\sqrt{14}\times\sqrt{17}} 0.908
5. Angle θ (in degrees) \cos^{-1}(0.908) 24.7°

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