Vector A has a magnitude of 10.6 units and points due west. Vector & points due north. (a) What is the magnitude of & if A + & has a

magnitude of 19.6 units? (b) What is the direction of A + & relative to due west? (c) what is the magnitude of & if A - & has a

magnitude of 19.6 units? (d) What is the direction of A - & relative to due west?

(Note: Use the convention that an angle north of west is positive, and an angle south of west is

Let's solve this problem step by step.

### Given:
- Vector **A** has a magnitude of 10.6 units and points due west.
- Vector **B** points due north.
- The magnitude of the vector sum **A + B** is 19.6 units.

### Part (a): Magnitude of **B** if **A + B** has a magnitude of 19.6 units

We know that **A** and **B** are perpendicular to each other because **A** points west and **B** points north. Therefore, we can use the Pythagorean theorem to find the magnitude of **B**:

\[
\text{Magnitude of } A + B = \sqrt{A^2 + B^2}
\]

Substituting the given values:

\[
19.6 = \sqrt{(10.6)^2 + B^2}
\]

Now, solving for **B**:

\[
19.6^2 = (10.6)^2 + B^2
\]
\[
384.16 = 112.36 + B^2
\]
\[
B^2 = 384.16 - 112.36
\]
\[
B^2 = 271.8
\]
\[
B = \sqrt{271.8} \approx 16.48 \text{ units}
\]

So, the magnitude of **B** is approximately **16.48 units**.

### Part (b): Direction of **A + B** relative to due west

To find the direction of the vector **A + B** relative to due west, we can use the tangent function:

\[
\tan(\theta) = \frac{B}{A} = \frac{16.48}{10.6}
\]

Calculating the angle **θ**:

\[
\theta = \tan^{-1}\left(\frac{16.48}{10.6}\right) \approx \tan^{-1}(1.554) \approx 57.5^\circ \text{ north of west}
\]

So, the direction of **A + B** relative to due west is approximately **57.5° north of west**.

### Part (c): Magnitude of **B** if **A - B** has a magnitude of 19.6 units

For the vector difference **A - B**, we again use the Pythagorean theorem:

\[
\text{Magnitude of } A - B = \sqrt{A^2 + B^2}
\]

Since **A - B** is also perpendicular (but pointing in different directions), the magnitude is the same as **A + B**:

\[
19.6 = \sqrt{(10.6)^2 + B^2}
\]

We already solved this in part (a), so the magnitude of **B** is again **16.48 units**.

### Part (d): Direction of **A - B** relative to due west

For the direction of **A - B**, we calculate:

\[
\tan(\theta) = \frac{B}{A} = \frac{16.48}{10.6}
\]

But this time, the direction is **south of west** since **B** is subtracted:

\[
\theta = \tan^{-1}\left(\frac{16.48}{10.6}\right) \approx 57.5^\circ \text{ south of west}
\]

So, the direction of **A - B** relative to due west is approximately **57.5° south of west**.

### Summary of Results:
(a) Magnitude of **B** = **16.48 units**

(b) Direction of **A + B** = **57.5° north of west**

(c) Magnitude of **B** (for **A - B**) = **16.48 units**

(d) Direction of **A - B** = **57.5° south of west**

Would you like any further details or explanations on these results?

Here are some related questions for further exploration:
1. How would the results change if vector **B** pointed east instead of north?
2. What would be the magnitude and direction of **A - 2B**?
3. How do these vector operations apply to real-world navigation?
4. Can you calculate the resultant vector if **B** had a different magnitude?
5. How does changing the angle between **A** and **B** affect the results?

**Tip:** When working with vector problems, always visualize the vectors on a coordinate system to help understand their directions and relationships better.

Explore vector addition and subtraction problems involving magnitude and direction. Solve for the magnitude and direction of vectors A + B and A - B using the Pythagorean theorem and trigonometric functions. Suitable for high school students studying vectors.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation