Question

Vector a has a magnitude of 4 and points 80° north of east. Vector b has a magnitude of 10 and points 10° south of east. Find a + b . Include the direction

Answer 1

To find the vector sum \( \mathbf{a} + \mathbf{b} \), we first need to break down both vectors into their components.

### Step 1: Components of Vector \( \mathbf{a} \)

Vector \( \mathbf{a} \) has a magnitude of 4 and points 80° north of east.

- **East (x-axis)** component of \( \mathbf{a} \):
\[
a_x = 4 \cos(80^\circ)
\]
- **North (y-axis)** component of \( \mathbf{a} \):
\[
a_y = 4 \sin(80^\circ)
\]

Calculating these:
\[
a_x = 4 \cos(80^\circ) \approx 4 \times 0.1736 \approx 0.6944
\]
\[
a_y = 4 \sin(80^\circ) \approx 4 \times 0.9848 \approx 3.9392
\]

Thus, the components of vector \( \mathbf{a} \) are:
\[
\mathbf{a} = (0.6944, 3.9392)
\]

### Step 2: Components of Vector \( \mathbf{b} \)

Vector \( \mathbf{b} \) has a magnitude of 10 and points 10° south of east.

- **East (x-axis)** component of \( \mathbf{b} \):
\[
b_x = 10 \cos(10^\circ)
\]
- **South (negative y-axis)** component of \( \mathbf{b} \):
\[
b_y = -10 \sin(10^\circ)
\]

Calculating these:
\[
b_x = 10 \cos(10^\circ) \approx 10 \times 0.9848 \approx 9.848
\]
\[
b_y = -10 \sin(10^\circ) \approx -10 \times 0.1736 \approx -1.736
\]

Thus, the components of vector \( \mathbf{b} \) are:
\[
\mathbf{b} = (9.848, -1.736)
\]

### Step 3: Sum of the Vectors \( \mathbf{a} + \mathbf{b} \)

Now we can find the components of the resultant vector \( \mathbf{c} = \mathbf{a} + \mathbf{b} \):

- **x-component of \( \mathbf{c} \)**:
\[
c_x = a_x + b_x = 0.6944 + 9.848 \approx 10.5424
\]
- **y-component of \( \mathbf{c} \)**:
\[
c_y = a_y + b_y = 3.9392 - 1.736 \approx 2.2032
\]

So, the resultant vector \( \mathbf{c} \) is:
\[
\mathbf{c} = (10.5424, 2.2032)
\]

### Step 4: Magnitude and Direction of \( \mathbf{c} \)

**Magnitude of \( \mathbf{c} \)**:
\[
|\mathbf{c}| = \sqrt{c_x^2 + c_y^2} = \sqrt{(10.5424)^2 + (2.2032)^2} \approx \sqrt{111.0176 + 4.8511} \approx \sqrt{115.8687} \approx 10.77
\]

**Direction of \( \mathbf{c} \)**:
The direction \( \theta \) relative to east can be calculated using the tangent function:
\[
\tan(\theta) = \frac{c_y}{c_x} = \frac{2.2032}{10.5424}
\]
Calculating \( \theta \):
\[
\theta \approx \tan^{-1}\left(\frac{2.2032}{10.5424}\right) \approx \tan^{-1}(0.208) \approx 11.73^\circ
\]

### Final Result

The resultant vector \( \mathbf{c} \) can be stated as:
\[
\mathbf{c} \approx 10.77 \text{ units at } 11.73^\circ \text{ north of east}.
\]