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------------------------------------------------ Given the origin-based vector \( \mathbf{A} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \), what is the angle between \(\mathbf{A}\) and the x-axis?

A. 22 Degrees
B. 24 Degrees
C. 66 Degrees
D. 80 Degrees

To find the angle between vector A = i + 2j + k and the x-axis, we can use the formula cos(θ) = Ax / |A|. Substituting the values, we find that θ ≈ 43.6 degrees.

Explanation:

To find the angle between vector A = i + 2j + k and the x-axis, we need to find the direction angle of the vector A. The direction angle is the angle made by the vector with the positive x-axis. Since the vector A has a component i in the x-direction, the direction angle is equal to the angle made by the vector A with the x-axis. We can use the formula:

cos(θ) = Ax / |A|

where Ax is the x-component of vector A and |A| is the magnitude of vector A. In this case, Ax = 1 and |A| = √(12 + 22 + 12) = √6. Substituting these values into the formula, we have:

cos(θ) = 1 / √6

θ ≈ 43.6 degrees