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.