SOLUTION: Find the magnitude and direction of the resultant for Vq = 5 ms and Vp 6ms. with angle 120 degrees between the two lines.

Question 1178658: Find the magnitude and direction of the resultant for Vq = 5 ms and Vp 6ms. with angle 120 degrees between the two lines.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down how to find the magnitude and direction of the resultant vector.
**1. Represent the Vectors:**
* **Vq:** Magnitude = 5 ms, Direction = 0 degrees (we can assume it's along the x-axis for simplicity).
* Vq = 5i + 0j
* **Vp:** Magnitude = 6 ms, Direction = 120 degrees.
* Vp_x = 6 * cos(120°) = 6 * (-1/2) = -3 ms
* Vp_y = 6 * sin(120°) = 6 * (√3 / 2) = 3√3 ms
* Vp = -3i + 3√3j
**2. Find the Resultant Vector (Vr):**
* Vr = Vq + Vp
* Vr = (5i + 0j) + (-3i + 3√3j)
* Vr = (5 - 3)i + (0 + 3√3)j
* Vr = 2i + 3√3j
**3. Find the Magnitude of the Resultant (||Vr||):**
* ||Vr|| = √(Vr_x² + Vr_y²)
* ||Vr|| = √(2² + (3√3)²)
* ||Vr|| = √(4 + 27)
* ||Vr|| = √31
* ||Vr|| ≈ 5.57 ms
**4. Find the Direction of the Resultant (θ):**
* θ = arctan(Vr_y / Vr_x)
* θ = arctan(3√3 / 2)
* θ = arctan(5.196 / 2)
* θ = arctan(2.598)
* θ ≈ 68.96 degrees
**Therefore:**
* **Magnitude of the Resultant:** √31 ms (approximately 5.57 ms)
* **Direction of the Resultant:** Approximately 68.96 degrees (measured counter-clockwise from the x-axis, or the direction of Vq).

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