V1 x-component: 2.46
V1 y-component: 0.43
V1 x-direction: W
V1 y-direction: N
V2 x-component: 1.93
V2 y-component: 2.30
V2 x-direction: E
V2 y-direction: N solve the vector magnitude and angle

To solve for the resultant magnitude and angle of the vectors \(V_1\) and \(V_2\), we first need to determine their components in Cartesian form and then combine them.

### Step 1: Determine the Cartesian components of the vectors.

#### For \(V_1\):
- \(V_1\) x-component = 2.46 (directed West, hence negative in Cartesian form: -2.46)
- \(V_1\) y-component = 0.43 (directed North, positive in Cartesian form: +0.43)

So the Cartesian form of \(V_1\) is:
\[
V_1 = (-2.46, +0.43)
\]

#### For \(V_2\):
- \(V_2\) x-component = 1.93 (directed East, positive in Cartesian form: +1.93)
- \(V_2\) y-component = 2.30 (directed North, positive in Cartesian form: +2.30)

So the Cartesian form of \(V_2\) is:
\[
V_2 = (+1.93, +2.30)
\]

### Step 2: Add the components of \(V_1\) and \(V_2\).

#### Resultant x-component:
\[
R_x = (-2.46) + (+1.93) = -0.53
\]

#### Resultant y-component:
\[
R_y = (+0.43) + (+2.30) = +2.73
\]

### Step 3: Calculate the magnitude of the resultant vector.

The magnitude \(R\) of the resultant vector is given by the Pythagorean theorem:
\[
R = \sqrt{R_x^2 + R_y^2}
\]

Substitute the values:
\[
R = \sqrt{(-0.53)^2 + (2.73)^2} = \sqrt{0.2809 + 7.4529} = \sqrt{7.7338}
\]

\[
R \approx 2.78
\]

### Step 4: Calculate the direction (angle) of the resultant vector.

The angle \(\theta\) with respect to the positive x-axis is given by:
\[
\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]

Substitute the values:
\[
\theta = \tan^{-1}\left(\frac{2.73}{-0.53}\right)
\]

\[
\theta \approx \tan^{-1}(-5.15) \approx -79^\circ
\]

Since the resultant vector lies in the second quadrant (negative x, positive y), the actual angle is:
\[
\theta = 180^\circ - 79^\circ = 101^\circ
\]

### Final Answer:
- Magnitude of the resultant vector \(R \approx 2.78\)
- Direction of the resultant vector \( \theta \approx 101^\circ \) (measured counterclockwise from the positive x-axis)

Would you like more details or clarifications on any step?

### 5 Related Questions:
1. How would the result change if the second vector had a negative y-component?
2. Can you calculate the unit vector of the resultant vector?
3. How can we decompose the resultant vector back into its x- and y-components geometrically?
4. How would the vector calculation change in 3D space?
5. Can you explain how vectors are represented in polar coordinates?

### Tip:
Always be mindful of the quadrant when determining angles, as the arctangent function might return a negative angle that needs adjustment based on the vector's components.

Learn how to solve for the magnitude and direction of two 2D vectors with components V1 and V2. This solution involves vector addition, Pythagorean theorem, and trigonometric functions. Suitable for students in Grades 10-12.

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