There are 3 vectors in a 2D coordinate with unit vector 𝑥̂ and 𝑦̂ along x and y
axes respectively. The magnitude of 𝑎⃗ is 5 and it forms an angle of 60𝑜 with the positive
x-axis. Vector 𝑏⃗⃗ = 5𝑥̂ + 8𝑦̂. The coordinates of the starting and ending points of 𝑐⃗ are (4,
5) and (12, 7). Calculate the resultant of 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗

Let's break this problem down step by step, starting with calculating each vector in terms of its components.

### Step 1: Vector \( \mathbf{a} \)
- The magnitude of \( \mathbf{a} \) is given as 5, and it forms an angle of \( 60^\circ \) with the positive x-axis.
- Using trigonometry, we can find the components of \( \mathbf{a} \) along the \( \hat{x} \) and \( \hat{y} \) directions.

\[
\mathbf{a_x} = 5 \cdot \cos(60^\circ) = 5 \cdot \frac{1}{2} = 2.5
\]
\[
\mathbf{a_y} = 5 \cdot \sin(60^\circ) = 5 \cdot \frac{\sqrt{3}}{2} \approx 4.33
\]
So, vector \( \mathbf{a} \) can be expressed as:
\[
\mathbf{a} = 2.5 \hat{x} + 4.33 \hat{y}
\]

### Step 2: Vector \( \mathbf{b} \)
- Vector \( \mathbf{b} \) is directly given as \( \mathbf{b} = 5 \hat{x} + 8 \hat{y} \).

### Step 3: Vector \( \mathbf{c} \)
- The coordinates of the starting point of vector \( \mathbf{c} \) are \( (4, 5) \) and the ending point is \( (12, 7) \).
- To find \( \mathbf{c} \), we calculate the difference between the ending and starting coordinates.

\[
\mathbf{c_x} = 12 - 4 = 8
\]
\[
\mathbf{c_y} = 7 - 5 = 2
\]
Thus, vector \( \mathbf{c} \) can be expressed as:
\[
\mathbf{c} = 8 \hat{x} + 2 \hat{y}
\]

### Step 4: Sum of the vectors \( \mathbf{a} + \mathbf{b} + \mathbf{c} \)
Now, let's sum the three vectors component-wise.

#### x-component:
\[
\mathbf{R_x} = \mathbf{a_x} + \mathbf{b_x} + \mathbf{c_x} = 2.5 + 5 + 8 = 15.5
\]

#### y-component:
\[
\mathbf{R_y} = \mathbf{a_y} + \mathbf{b_y} + \mathbf{c_y} = 4.33 + 8 + 2 = 14.33
\]

### Step 5: Magnitude of the resultant vector
The magnitude \( R \) of the resultant vector can be found using the Pythagorean theorem:
\[
R = \sqrt{(\mathbf{R_x})^2 + (\mathbf{R_y})^2} = \sqrt{(15.5)^2 + (14.33)^2}
\]
\[
R \approx \sqrt{240.25 + 205.35} = \sqrt{445.6} \approx 21.11
\]

### Step 6: Direction of the resultant vector
The direction \( \theta \) (with respect to the positive x-axis) can be found using the tangent function:
\[
\theta = \tan^{-1} \left( \frac{\mathbf{R_y}}{\mathbf{R_x}} \right) = \tan^{-1} \left( \frac{14.33}{15.5} \right)
\]
\[
\theta \approx \tan^{-1} (0.924) \approx 42.67^\circ
\]

### Final Result:
The resultant vector \( \mathbf{R} \) has:
- **Magnitude**: approximately \( 21.11 \)
- **Direction**: approximately \( 42.67^\circ \) above the positive x-axis.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the result change if the angle of vector \( \mathbf{a} \) was different?
2. Can you verify the magnitude and direction of the resultant vector using graphical methods?
3. How does the position of vector \( \mathbf{c} \) affect the overall resultant?
4. What would happen if vector \( \mathbf{b} \) was negative in both components?
5. How can you decompose the resultant vector into unit vector form?

**Tip**: Always keep track of the angle direction (above or below the x-axis) when calculating vector components!

There are 3 vectors in a 2D coordinate with unit vector 𝑥̂ and 𝑦̂ along x and y axes respectively. The magnitude of 𝑎⃗ is 5 and it forms an angle of 60° with the positive x-axis. Vector 𝑏⃗⃗ = 5𝑥̂ + 8𝑦̂. The coordinates of the starting and ending points of 𝑐⃗ are (4, 5) and (12, 7). Calculate the resultant of 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗.

Learn how to calculate the resultant vector of three 2D vectors: a vector with magnitude 5 forming a 60° angle, a given vector b = 5x + 8y, and a vector c from coordinates (4, 5) to (12, 7). This problem explores vector addition, trigonometry, and coordinate geometry for students in Grades 10-12.

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