Determining the Direction Angle Between a Force Vector and the x-Coordinate Axis

What is the direction angle between a force vector of magnitude 450N directed from C(-3,4,0) to D(1,5,3) and the x-coordinate axis?

Calculating the Direction Angle:

To determine the direction angle between the force vector and the x-coordinate axis, we can use the dot product formula: cos θ = (Q dot i) / |Q| * |i| Where: - Q is the force vector - i is the unit vector along the x-axis To calculate Q dot i, we find the dot product between the force vector Q and the unit vector i: Q dot i = Qx * ix + Qy * iy + Qz * iz Given components: - Qx = 1 - (-3) = 4 - Qy = 5 - 4 = 1 - Qz = 3 - 0 = 3 The unit vector i has components: - ix = 1 - iy = 0 - iz = 0 Substitute the values into the dot product formula: Q dot i = 4 * 1 + 1 * 0 + 3 * 0 = 4 Next, calculate the magnitude of Q: |Q| = √(Qx² + Qy² + Qz²) = √(4² + 1² + 3²) = √(16 + 1 + 9) = √26 Magnitude of i: |i| = √(ix² + iy² + iz²) = √(1² + 0² + 0²) = √1 Substitute values into the direction angle formula: cos θ = 4 / (√26 * √1) = 4 / √26 Find the angle θ: θ = cos⁻¹(4 / √26) ≈ 56.93 degrees Therefore, the direction angle between the force vector and the x-coordinate axis is approximately 56.93 degrees.

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