Arrange the vectors subtractions so that their magnitudes are in decreasing order. If the two vectors `vecA` and `vecB` are acting at an angle `(|vecA| gt |vecB|)`
a) 60° b) 90° c) 180° d) 120°

Options

A. d,c,b,a
B. a,b,d,c
C. c,d,b,a
D. c,d,a,b

Answer

To solve the problem of arranging the magnitudes of vector subtractions in decreasing order based on the angles between vectors \(\vec{A}\) and \(\vec{B}\), we will use the formula for the magnitude of the resultant vector when two vectors are subtracted. The formula is given by:

\[ |\vec{S}| = |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta} \]

where \(\theta\) is the angle between the two vectors.

### Step 1: Calculate the Magnitude for Each Angle

1. **For \(\theta = 60^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(60^\circ)} \]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - |\vec{A}| |\vec{B}|} \]

2. **For \(\theta = 90^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(90^\circ)} \]
Since \(\cos(90^\circ) = 0\):
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2} \]

3. **For \(\theta = 120^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(120^\circ)} \]
Since \(\cos(120^\circ) = -\frac{1}{2}\):
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + |\vec{A}| |\vec{B}|} \]

4. **For \(\theta = 180^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(180^\circ)} \]
Since \(\cos(180^\circ) = -1\):
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}|} \]

### Step 2: Compare the Magnitudes

Now we will summarize the results for each angle:

- **For \(60^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - |\vec{A}| |\vec{B}|} \]

- **For \(90^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2} \]

- **For \(120^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + |\vec{A}| |\vec{B}|} \]

- **For \(180^\circ\)**:
\[ |\vec{S}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}|} \]

### Step 3: Order the Magnitudes

To arrange in decreasing order, we observe the following:

- The magnitude for \(180^\circ\) will be the largest because it adds the most positive terms.
- The magnitude for \(120^\circ\) is next, as it also adds a positive term.
- The magnitude for \(90^\circ\) is next, as it has no subtraction.
- The magnitude for \(60^\circ\) is the smallest due to the subtraction of the term.

Thus, the order of magnitudes from greatest to least is:

1. \(180^\circ\)
2. \(120^\circ\)
3. \(90^\circ\)
4. \(60^\circ\)

### Final Answer

The arrangement of the vectors in decreasing order of their magnitudes is:
**c) 180°, d) 120°, b) 90°, a) 60°**